ALGEBRA FOR BEGINNERS.

WITH NUMEROUS EXAMPLES .

AT QADHUNTER, M.A. ,F.R .S .

I

NEW EDITION .

flonhon

MACMILLAN AND 00 .

floronto

COPP, CLARK AND 00 .

1 87 5 .

[All Righ ts r eeervedj

PREFA CE .

THE presen t work h as been undertaken at the requestof many teachers, in order to be placed in the hands ofbeginn ers, and to serve as an introduction to the largertreatise published by the author ; it is accordingly basedo n the earlier chapters o f that treatise, but is o f a moreelemen tary character. G reat pain s have been taken torender the work in telligible to young studen ts, by the useo f simple language and by copious explan ation s.

In determin ing the subjects to be in cluded and th e

space to be assign ed to each, the author h as been guidedby the papers given at the various examination s in e le

men tary Algebra which are n ow carried on in this coun try.

Th e book may be said to con sist o f three parts. The firstpart con tain s the elemen tary operation s in in tegral and

fraction al expression s ; it occupies eighteen chapters. Th e

second part con tain s the solution of equations and proh lems ; it occupies twelve chapters. The subjects con tainedin these two parts con stitute n early the whole o f every ex

amination paper which was con sulted,and accordingly they

ar e treated with ample detail o f illustration and exercise .The third part forms th e remainder o f the book ; it consists o f vari ous subj ects which ar e in troduced but rarelyin to the examination papers

,and which ar e therefore more

briefly discussed.

Th e subjects are arranged in what appears to be themost natural order. But many teachers find it advan

tageons to in troduce easy equation s and problems at a veryearly stage, and accordingly provision h as been made for

PREFA CE .

THE presen t work h as been undertaken at the requestof many teachers, in order to be placed in the hands o f

beginn ers, and to serve as an in troduction to the largertreatise published by th e author ; it is accordingly basedo n the ear lier chapters of that treatise, but is of a moreelemen tary character. G reat pain s have been taken torender the work in telligible to young studen ts, by the useof simple language and by copious explan ation s.In determin ing the subjects to be in cluded and th e

space to be assign ed to each, the author h as been guidedby the papers given at the various examination s in e le

men tary Algebra which are n ow carried on in this coun try.

Th e book may be said to con sist o f three parts. The firstpart con tain s the elemen tary operation s in in tegral an d

fraction al expression s ; it occupies eighteen chapters. Th e

second part con tains the solution of equations and problems ; it occupies twelve chapters. Th e subjects con tainedin these two parts con stitute n early the whole o f every ex

amination paper which was con sulted, and accordingly theyar e treated with ample detail o f illustration and exercise .

The third part forms th e remainder o f the book ; it consists o f various subjects which ar e in troduced but rarelyin to the examination papers

,and which ar e therefore more

briefly discussed.

Th e subjects are arranged in what appears to be themost n atural order. But many teachers find it advan

tage ous to in troduce easy equation s an d problems at a veryearly stage, and accordingly provision has been made for

vi PREFACE .

such a course . It will be found that Chapters XIX. and

XXI. may be taken as soon as a studen t has proceeded as

far as algebraical multipl ication .

In accordan ce with the recommendation of teachers, theexamples for exercise are very numerous . Some o f thesehave been selected from the College and Un iversity examin ation papers

,and some from the works of Saunderson and

Simpson ; many however are or iginal, and ar e con structedwith referen ce to poin ts which have been shewn to be importan t by the author

’

s experience as a teacher and an

examin er.Th e author h as to ackn owledge the kindness o f many

distinguished teachers who have examin ed the sheets o f hiswork and have given h im valuable suggestions . Any r e

marks on the work, an d especially the in dication of didicul ties either in the text or the examples, will be mostthankful ly received.

I . TODHUNTER.

ST JOHN’S COLLEGE

, A!P / _

July 1 863 .

0

Four n ew Chapters have been added to the presen t edition

,and also a collection of Miscellan eous Examples which

ar e arranged in se ts,each set con taining te n examples.

These addition s have been made at the request o f someemin en t teachers, in order to increase the utili ty of thework.

CONTENTS.

PAGE

1 Th e Prin cipal Sign sII . Factor. Coefficien t. Power. Terms

III . R emain ing Sign s . BracketsIV. Ch an ge o f the order of Terms . Like Terms

V . Addition ’

.

G en eral Resul ts in M ultiplicationFacto rsG reatest Common M eas ur e

Least Common M ultipleFractionsReductio n o f Fr actio n sAdditio n or Subtraction of Fraction sM ul tiplication o f Fractio n sD ivisio n o f Fraction sSimple Equatio n sSimple Equation s, con tin uedProblems

Problems, con tin ued

Simultan eous Equatio n s o f th e first degreewith two unkn own quantities

XX IV . Simultan eous Equations of th e fir st degreewith m ore th an two unkn own

XXV. Problems which lead t o simultan eous equa

tio n s o f th e first degree with m ore thano n e unkn own quan tity

vi PREFACE .

such a course. It will be found that Chapters XIX. and :

XXI. may be taken as soon as a studen t has proceeded asfar as algebraical multiplication .

In accordan ce with the recommendation of teachers, theexamples for exercise are very numerous . Some o f thesehave been selected from the Coll ege and Un iversity examin ation papers

,an d some from the works o f Saunderson and

Simpson ; many however ar e original, and ar e con structedwith referen ce to points which have been shewn to be importan t by the author

’

s experience as a teacher and an

examiner.Th e author h as to acknowledge the kindness o f many

distinguished teachers who have examin ed the sheets o f hiswork and have given h im valuable suggestion s. Any r e

marks on the work, and especially the in dication o f dithculties either in the text or the examples

,will be most

thankfully received.

I. TODHUNTER.

ST JOHN’S COLLEGE

, A};

llJuly 1 863 . ; 4 ¢r

I

Four n ew Chapters have been added to the presen t edition

, and also a collection of M iscellan eous Examples whichare arranged in se ts

,each se t con tain ing ten examples.

These additions have been made at'

the request of some“emin en t teachers

,in order to increase the utility o f the

work.

July 1 867.

CONTENTS.

Factor. Coeffi cien t. Power. Term s

R emain ing Sign s . BracketsChan ge of the order o f Terms. Like Terms

Addition .

SubtractionBrackets .

M ultiplication

G en eral Resul ts in M ultiplicationFacto rsGreatest Common M eas ur e

Leas t Comm on M ultipleFractio nsReductio n o f Fraction sA dditio n or Subtraction of FractionsM ultiplication o f Fraction sD ivisio n o f Fraction sS imple Equatio n sSimpl e Equation s, con tin uedProblems

Pr oblem s, con tin ued

Simultan eous Equation s o f th e first degreewith two unkn own quan tities a “

XX IV . Simultan eous Equations o f th e first degre ewith more than two unkn own

Problem s which lead to simultan eous equa

tio n s of th e first degre e with m ore thano n e un kn own quan tity

XXVI . Quadratic Equatio nsXXVII. Equatio n s wh ich may be solved like Quad

ratics 1 7 1

X XV III . Problems which lead to Quadratic Equatio n s 1 76

XX IX . Simultan eous Equatio n s in volving Quad

XXX . Problems which l ead to Quadratic Equation swith more than o n e unkn own quan tityI n vo lutio n

Propor tionVariationArithmetical Progressio nG eometrical ProgressionHarmon ical Progressio nPe rmutatio n s and Combin atio nsBin omial TheoremScal es o f NotationInterestM iscellan eous Examples

ANSWERS

ALGEBRA FOR BEGINNERS.

I. Th e Pr in cipal S ign s.

1 . A LGEBRA is the scien ce in which we reason aboutnumbers, with the aid o f letters to den ote the numbers

,

and o f certain signs to den ote the Operation s performedo n the numbers

,and the relations of the mnn bers to each

other.

2. Numbers may be either kn own numbers,o r num

bers which have to be found, an d which ar e thereforecalled un kn own n umbers. It is usual to represen t kn ownnumbers by the first letters Of the alphabet, a

, b, e , &c.,

and un known numbers by the last letters x, y ,

z ; this ishowever n o t a n ecessary rule, an d so n eed n o t be strictlyobeyed . Numbers may be either whole o r fraction al . Theword qua n tity is Often used with the same mean ing as

n umber . The word in teger is often used in stead o f wh olen umber .

3 . Th e beginn er h as to accustom himself to the use Ofletters for represen ting numbers, an d to learn the mean ingo f the signs ; we shall begin by explain ing the most importan t sign s an d illustrating their use . We shall assumethat the studen t h as a kn owledge Of the elemen ts Of Arithmetic

,an d that he admits the truth o f the common n otion s

required in all parts o f mathematics , such as, if equa ls beadded to equals th e wh oles ar e equal , and the like.

4 . Th e sign + placed before a number den otes that thenumber is to be added. Thus a b den otes that the number represen ted by b is to be added to the number r epre

l

2 THE PRINCIPAL SIGNS .

sen ted by a . If a represen t 9 and b represen t 3 , then a brepresen ts 12. The sign is call ed the p lus sign , and

a + b is read thus “a p lus b.

”

5 . Th e sign —placed before a number den otes that thenumber is to be subtr ac ted. Thus a —b den otes that thenumber represen ted by b is to be subtrac ted from thenumber represen ted by a . If a represen t 9 an d b r epre

sen t 3,then a b represen ts 6 . The sign is called the

m in us s ign , and a — b is read thus a.m in us b.

”

6 . Similarly a + b+ e den otes that we are to add b toa,and then add 0 to the resul t ; u + b— c den otes that we

ar e to add b to a,and then subtract 0 from the resul t ;

a -b+ 0 den otes that we ar e to subtract b from a,and then

add 0 to the result ; u—b - c den otes that we ar e to sub

tract b from a,and then subtract 0 from the resul t.

7. The sign den otes that the numbers betweenwhich it is placed are equa l. Thus u z b den otes that thenumber represen ted by a is equal to the number r epr esen ted by b. And a + b= e den otes that the sum Of thenumbers represen ted by a and b is equal to the numberrepresen ted by 0 ; so that if a represen t 9 , and b repres en t3 , then 0 must represen t 12. The sign is call ed thesign of equal ity , and u z b is read thus a equa ls b or“a is equal to b.

”

.

8 . Th e sign x den otes that the numbers betweenwhich it stands ar e to be multip l ied together. Thusa x b

.

den otes that the n umber represen ted by a is to bemul tiplied by the n umber represen ted by b. If a r epr esen t 9, and b represen t 3 , then a x b represen ts 27. Th eS ign x is called the S ign of m ult ip lica tion , and a x b is

read thus “a in to b.

”Similarly a x b x 0 den otes the pro~

duct of the n umbers represen ted by a , b, and c.

9 . Th e sign of multiplication is however ofte n omittedfor the sake of brevity ; thus ab is used in stead o f a x b,and has the same mean ing ; so also abe is used ins tead o fa x b x c

,and h as the same mean ing.

Th e sign o f multiplication mus t n o t be omitted whennumbers ar e expressed in the ordinary way by fig uresThus 4 5 cann ot be used to represen t the product Of 4 and

THE PRINCIPAL SIGNS . 3

5,becaus e a differen t mean ing h as already been appro

priated to 4 5 , n amely, f or ty -fine. We must therefore r e

presen t the product Of 4 an d 5 in an other way, an d 4 x 513 the way which is adopted. Sometimes

,however

, a

poin t is used ins tead of the sign x thus is used ihstead Of 4 x 5 . To preven t any con fusion between thepoin t thus us ed as a sign o f multiplication

,an d the poin t

used in the n otation for decimal fraction s,it is advisable

to place the poin t in the latter case h igher up ; thus5

4 5 may be kept to den ote 4 +175

But in fact the poin t isn o t us ed in stead Of the sign x except in case s where therecan be n o ambiguity. Fo r example, may be put for1 x 2 x 3 x 4 because the poin ts here will n o t be taken fordecimal poin ts.

Th e poin t is sometimes placed instead o f the sign x

between two letters ; so that a . b is used ins tead of a x b.

But the poin t is here superfluous,because

,as we have

said,ab is used in stead o f a x b. No r is the poin t

,n o r th e

sign x n ecessary between a number expressed in the oudin ary way by a figu re and a number represen ted by a

letter ; so that,for example

, 3 a is used in stead of 3 x a,and has the same mean ing.

10. Th e sign den ote s hat the number whi ch pre~cedes it is to be divided by the number which follows it.Thus a + b den otes that the number represen ted by a is tobe divided by the number represen ted by b. If a repre

sent 8, and b represen t 4 , then u+ b represen ts 2. Thesign is called the sign of division ,

and a—z—b is readthus “

a by b .

”

There is also an other way of den oting that o n e numh er is to be divided by an other ; the dividend is placed

over the divisor with a lin e between them. Thus is

used in stead Of a—z—b,and h as the same meaning.

1 1 . Th e letters of the alphabet,

“

and . the signs whichwe have already explain ed, together with those which mayoccur hereafter

,ar e called algebr a ical symbols, becaus e

they ar e used to repres en t the numbers about which wemay be reasoning, the Operations performed on them and

4 EXAMPLES . I .

their relations to each other. Any collection of Algebraicalsymbols is called an a lgebr a ica l exp r ession , or briefly an

12. We shall n ow give some examples as an exercis ein the use o f the symbols which have been explain ed ;these examples con sist in fin ding th e numerical values o f

certain algebraical expressions .

Suppose a = 1 , b= 2, c= 3 , d -é 5,

Then7a + 3 b

2ab+ 8bc

4m: l Obe de 12 120 3 0= 1

b+

cd ac 24—1 5 3

10 4 10 4.

4 c + 5 c 12+ 3o

d - b 5 —2 3

EXAMPLES. I.

If a = 1,b= 2, c : 3 , d 4

,e : 5

, f : 0,find the numeri

cal values o f the following expressions .

1. 9a + 2b+ 3 c — 2jf 2. 4 e—3 a - 3b+ 5 e.

3 . 7ae 3 bc 9d af . 4 . Saba bcd+ 9cde dry:

5 . abcd abce abde acde bode. 6 .

4 a 99 80 5 4

b e d e

4ao . 8b0 5 cd 120+Gb 200

“

b d e be c7 t+de

'

cde 5 bcd (Sade

ab a e be10° 76 4—d

2a + 5 b 3 b+ 2e a + b+ c + d

c+

d 2c

https://www.forgottenbooks.com/join

6 FACTOR . COEFFICIEN T. P0WER . TERM S .

1 6 . A power is more briefly den oted thus : in stead o f

expressing all the equal factors, we express the factor on ce,an d place over it the number which indicates how oft en itis to be repeated. Thus a2 is used to den ote a x a ; a

3 is

used to den ote a x a x a ; a4 is us ed to den ote a x a x a x a ;

and so on . An d a1 may be used to den ote the first power

o f a,that is a itself ; so that a 1 has the same mean ing as a .

17. A number placed over an other to indicate howmany times the latter occurs as a factor in a power

,is

called an index of th e p ower , or an exp on en t of th e p ower ;o r

,briefly, an in dex , o r exp on en t.Thus

,for example, in a

s the expon en t is 3 ; in a" the

expon en t is n .

18. Th e studen t must distinguish very carefully betweena coefi cien t and an exp on en t. Thus 30mean s th r ee times c

here 3 is a coefi cien t. But 03 means 0 times 0 times 0 ;here 3 is an exp on en t. That is

19 . Th e second power o f a , that is a2, is often called thesqua r e o f a , o r a squar ed ; and the third power o f a , that isa 3

,is often called the cube of a

, o r a cubed. There ar e n o

such words in use for the higher powers ; a 4 is read thus“a to th ef our th p ower , o r briefly a. to the f o ur th .

”

l

20. If an expression con tain n o par ts conn ected by thesign s and it is called a s imp le expression . If an

expression con tain parts conn ected by the sign s and

it is called a comp oun d expression , and t he parts con

n ected by th e signs and are called terms o f the expression .

Thus ax,4bc

,and 5 a2c2 ar e simple expression s ; a3 + b’ c

‘

is a compound express1on , and a”, b

"

,and c“are its terms.

21 . When an expression con sists o f two terms it iscalled a bin om ia l expression : when it con sists o f threeterms it is called a tr in omial expression ; any expressionconsisting o f several terms may be called a mul tin om ial

expression, or ap olyn omial expression .

FAaxon . COEFFICIEN T. E0WEB . TERM S . 7

Thus 2a + 3b is a bin omial expression ; a —2b+ 5 c is a

trin omial expression ; and a - b+ c d e may be called a.

multin omial expression o r a polyn omial expression .

22. Each o f the letters which occur in a term is

called a dimens ion o f the term,and the number o f the

letters is called the degr ee o f the term. Thus a7b3c - o r

a x a x bx b x b x c is said to be of six dimen sion s o r o f

the sixth degree. A numerical coefficien t is n o t coun ted ;thus 9a3b4 and a

3b4 are o f the same dimen sions, namelyseven dimen sion s. Thus the word d imen sion s refers tothe number o f algebraical multipl ication s involved in theterm ; that is, the degr ee o f a term, o r the n umber of itsdimen sion s , is the sum of th e exp on en ts of its a lgebr a ica l

factor s, provided we remember that if n o expon en t beexpressed the expon en t 1 must be understood, as indicatedin Art . 16 .

23 . An expression is said to be h om ogen eous when all itsterms ar e o f the same d imen sion s. Thus 7a3 + 3a2b 4abc

is homogen eous, for each term is of three dimen sion s.

We shall n ow give some more examples o f fin ding thenumerical values o f algebraical expression s.

Suppose a z l,b= 2, c= 3 , d : 4

,Then

b2= 4,b3 = 8, b4= 1 6, b5 z 3 2 .

3 5 2 12, 5 63 : 5 x 8 : 4 0

,288 .

eb= 5 2= 25

,e°= 5 3 = 125 .

a2b3= l x 8 : 8, 3b202= 3 x 4 x 9 : 108.

d3 + 02

3 02— 4 0— 1 0 27— 12 — 10 5

03 — 23 27— 18 + l 5 —23 1

c3 — a

3 27 — 1

e +d c— a 3 — 1

2 l - l 3 = 8.

8 EXA llIPLES . II.

EXAMPLES. II.

H a : I,b= 2 , c= 3 , d = 4 , find

values of th e following expressions :

e3 —d3 + c3 —b" + a

3

abc2+ bcdz dea2 +f3

03 — 2c

z+ 4c 13 .

e4+ 6e2b2+ b‘— 4 e

"”b— 4 eb

3

b202+de 32

4 a b" b"

2e + 2+3 c— 9

+e

‘ —1

e — 3 e— 2 e + 3°

a”

- l~ b2 024—5 3

e b

8a2 + 3b2 4 c2+ 6b2 b u d?

a2+ b” 0

2

4

czg+ bz+ c

a dz— C‘V—b2 a + e

‘ —d2'

a"+ 4a ib -1- 6a2b

‘

-1- b4

a3 3ab2 b3

REJIAINING’SIGNS BRAOKE TS . 9

III. Rema in ing S igns. B r ackets:

24 . Th e differen ce o f two numbers is sometimes den oted by the sign thus a ~ b den otes the differen ce o fthe n umbers represen ted by a and b ; and is equal to a — b,o r b a

, according as a is greater than b, o r less than b: butthis symbol is very rarely required.

25 . Th e sign den otes is gr ea ter th an , and the

sign den otes is less th an ; thus a >b den otes that thenumber represen ted by a is greater than the numberrepresen ted by b, and b< a den otes that the number represen ted by b is less than the numbe r represen ted by a .

Thus in both cases the open ing of the angle is turn edtowards the greater number.

26. Th e sign den otes th en o r th er ef or e ; the signden otes sin ce o r because.

27. Th e squar e r oot o f any assign ed number is thatnumber which has the assign ed number fo r its squar e orseco n d p ower . The cube r oot o f any assign ed number isthat number which has the assign ed number fo r its cube o rth ir d p ower . Th e f our th r oo t o f any assign ed n umber isthat number which h as the assign ed number for its four thpower. And so on .

Thus sin ce the square root of 4 9 is 7 ; and so ifa= b’, the square root of a is b. In like mann er, sin ce

the cube root o f 125 is 5 ; and so if a = o3, the cuberoot o f a is c.

28. Th e~

square root o f a may be den oted thus Ua ;but gen erally it is den oted simply thus Ja . The cube rootof a is den oted thus 2/a . The fourth root o f a is den otedthus i/a . And so on .

Thus J9= 3 3

Th e sign J is said to be a. corruption o f the in itialletter of the word r adix .

10 REMAININ G SIGNS . BRACKE r s.

29. When two o r more numbers are to be treated asforming o n e number they ar e en closed within br acke ts.

Thus , suppose we have to den ote that the sum of a and b

is to be mul tiplied by c ; we den ote it thus o r

{a + b} x c, or simply (a b) c o r {a b} 0 ; here we mean thatthe wh ole of a + b is to be multiplied by 0 . Now if we om itthe brackets we have a + bc, and this den otes that b on lyis t o be multiplied by c and the result added to a . Sim ilarly, (a + b

— c)d den otes that the result expressed bya + b— c is to be multiplied by d , o r that the wh ole ofa + b— c is to be multiplied by d ; but if we omit thebrackets we have a + b—cd

,and this den otes that 0 on ly

is to be multiplied by d and the result subtracted froma + b.

So also (a—b+ c) x (d e) den otes that th e result ex

pressed by a — b+ c is to be multiplied by the result expressed by d + e. This may also be den oted simply thus(a just as a x b is shorten ed in to ab.So also den otes that we ar e to obtain the

result expressed by a b+ 0,and then take the square roo t

o f this result.So also (ab)?den otes ab ab; and (ab)

sden otes ab ab x ab.

So also (a b—c) (d e) den otes that the resul t ex

Ipre

sised by a + b

—c is to be divided by the resul t expressedy e.

30. Sometimes in stead o f us ing brackets a lin e 18

drawn over the n umbers which ar e to be treated as formingo n e number. Thus a — b+ c x d + e is us ed with the samemeaning as (a —b+ c) x (d + e). A lin e used for this purpose 1s called a r in culum . So also (a + b—c) may

be den oted thus a + b— c

;d+ e

a + b— c and d e is really a c in culumused in a particularsen se.

and here th e lin e between

3 1 . We have n ow explain cd all the sign s which are

used in algebra. We may observe that in some cases theword s ign is appl ied specially to the two sign s and

thus in the Rule for Subtraction we shall speak of changing

EXAMPLES III . 1 1

th e signs, meaning the sign s and and in multiplication and division we shall speak o f theRu le of S ign s, meaning

/a rule relating to the signs and

3 2. We shall n ow give some more examples o f findingthe numerical values of expressions .

Suppose a = 1 , b= 2, c= 3 , d= 5 , e= 8. Then

«4 +

f/(4C

eJ (2b+ 4 c)—(2d — 2b)= 8 x 4 —8 x

J{(e—b)(20 J{(8 J(6 X

{(e - (d

f/(c3 3 c’b 3 cb2 b3)+ (a

2 bz 2ab)

J(1 + 4

EXAMPLES. III.

If a = 1, b= 2, c= 3 , d= 5 , e= 8,fin d the numerical

values o f the following exp ression s :

1 . a (b c). 2. b(c d) . 3 . c(c d)

b2(a’

62 5 . c

z(e2 b2 c

i). 6 .

961

223

52;62

. 8. J(3 bce). 9 . J(2b 4d 5 e).

(a + 2b+ 3c+ 5 e—4d)(6e—5 d—4 c—3b+ 2a).

(a2 b2+ d2 12. (3d

“37d

"

)2

eJ(d2 3c) dJ(d

23 c).

6 £76) — 4 )o

J(a2 2ab b”) x f/(a

33a2b 3ab

2 b3).

fj(c9—3 c"

'a 3ca

2a3) J(b

’+ c

z 2cb).

12 UIIANGE OF THE ORD ER OF TERflIS .

IV. Change of th e o rder of Terms. Like Terms.

3 3 . When all the terms o f an expression ar e conn ectedby the sign it is in differen t in what or der they are

placed ; thus and give the same result, n amely,12 ; and so also a + b and b+ a give the same result, n amely,th e sum o f the numbers which ar e represen ted by a and b.We may express this fact algebraically thus,

a + b= b+ a

Similarly,

3 4 . When an expression consists o f some terms preceded by the sign and some terms preceded by thesign we may wri te the former terms first in an y orderwe please, and the latter terms after them in any order weplease. This is obvious from the common n otions o f arithmetic. Thus

,for example

,

7+ s - 2—3 = s+ 7—2—3 = 7 + s—3 —3 - aa +b— c— e= b+ a—c—e= a + b— e—c= b+a—e— c.

3 5 . In some cases we may change the order o f theterms further, by mixing up the term s which are precededby the sign with those which ar e preceded by the signThus, for example, suppose that a represen ts 10, and b

'rekpresen ts 6, and 0 represen ts 5 , then

a + b—c= a —c + b= b - c+ a

for we arrive without any diffi culty at 1 1 as the result inall the cases.Suppose however that a represents 2, b represen ts 6,

and 0 represen ts 5,then the expression a—c + b presen ts a

difficulty, because we are thus apparen tly required to takea greater n umber from a less, n amely, 5 from 2. It willbe conven ien t to agree that such an expression as a—c + b,when 0 is greater than a

,shall be understood to mean th e

same thing as a + b— c. At presen t we shall n o t use suchan expression as a + b - c except when c is less than a b;

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14 LIKE TERMS .

like terms ; ai,5 a”

,and 9a2 are like terms ; a“, ab, and b2

ar e unhke terms.

40. An expression which con tain s like terms may besimplified. For example

,consider the expression

6a—a + 3b+ 5 c—b+ 3c —2a ;

by Art. 3 5 this expression is equ ivalen t to6 a—a - 2a + 3b—b 5 c+ 3c.

New 6a—a—2a= 3a ; for whatever number a may r e

presen t,if we subtract a from 6a we have 5 a left

,and then

if we subtra ct 2a from 5 a we have 3 a left. S imilar ly3b—b= 2b; and 5 0+ 3 c= 8a Thus the proposed expressionmay be put in the simpler form

3a 2b 8e.

Again ; consider the expression a — 3b—4b. This isequal to a — 7b. Fo r if we have first to subtract 3b froma n umber a , and then to subtract 4 b from the remain der,we shall obtain the required resul t in o n e Operation bysubtracting 7b from a ; this follows fr om the common n o

tion s o f Arithmetic. Thusa — 3b 4b= a—7b.

4 1 . There will be n o difliculty n ow in giving a mcan~

ing to such a statemen t as the following,—3b - 4b= —7b.

We cannot subtract 3b fr om n othing and then subtract4b from the remainder

,so that the statemen t just given is

n ot here in telligible in itself, separated from the rest o f analgebraical sen ten ce in which it may occur, but it can be

eas ily explain ed thus : if in the course o f an algebraicaloperation we have to subtract 3 b from a number and thento subtract 4 6 from the remainder

,we may subtra ct 7b at

on ce instead.

As the studen t advan ces in the subj ect he may be ledto conjecture that it is possible to give some mea ning tothe proposed statemen t by itsel f, that is , apart from anyother algebraical operation , and this conjec ture will befound correct, when a larger treatise 011 Algebra can be

EXAMPLES . IV. 15

consulted with advan tage ; but the explan ation which wehave given will be sufficien t fo r the presen t.

4 2. Th e simplifying o f expression s by coll ecting liketerms is the essen tial part o f the processes of Addition and

Subtraction in Algebra, as we shall se e in the n ext twoChapters.

It may be useful f01 the begin n er to n otice that acc01ding to o ur defin itions the following expressions me allequivalen t to the single symbol a :

a+ a

1, + 1 x a

,+ a x 1

,—1

EXAMPLES. IV.

If a = 1 , b= 2, c= 3,d = 4

,e= 5 , fi nd th e numerical

values of the f ollowing expression s :

a—3b+ 4 o . 2. a

4 a + 3 b 4 c + 3d+5 d + 4 e

b+ d a + d + a

(a —2b 3 c)”(b—2c+ 3d)

3+ (c—2d +

a“ 4a 3b Ga

2b2 4 ab3 V.

b2 2b0 c2

a“ 4 a3o 6a

2cz4ao3 0

4

2ab b2'

b“ 4 b30 6b202

4 bc3 + c“

7a —2b— 3 c - 4a + 5 b+ 4 0+ 2a .

5 a2+ 3 ab— 2b’ — ab 9b2—2ab 7b

2.

3 e 3—2a‘

3- i 5 a + a

3+ a + 9a

2—4a3 — 6aag+ 2ab b2 b2+ 2bc + c

2

a'

+ b

1 3 . JOE”5 412 e).

15 . i7<2e2+ 7b).

a .

02+ 20d+ d2

c + d

14 ; - ai).

1a UQN+8

16 AD D I TION .

V. Add ition .

4 3 . It is conven ien t to make three cases in Addition,

namely, I. When the terms are all like terms and have th esame sign ; II. When the terms ar e all like terms buthave n o t all the same sign ; III. When the te rms are n o tall like terms. We shall take these three cas es in order.

4 4. I. To add like te rms which have th e same sign.

Add th e n umer ical coefi cien ts, p r efix th e co mmon sign ,and an n ex th e common letter s.

For example, 6a 3 a 7a = l ca ,

—2bc 7bc 9bc= l o.

In the first example 6a is equivalen t to + 661, and 1 6ato + 16a. See Art . 3 8 .

4 5 . II. To add like terms which have n ot all thesame sign . Add al l th e p ositive n umer ical co efficien tsin to o n e sum

,an d a ll th e n ega tive n umer ica l coefficien ts

in to an other ; take th e dif er en ce of th ese two sum s,

p r efix th e sign of th e gr ea ter , and an n ex th e common.le tter s.

Fo r example,

7a—3a + 1 1a + a - 5 a — 2a = 19a — 10a= 9a,

2be—7be—3bc + 4bc + 5 bc— 6bc = l l bc—1 6bc= —sbe.

4 6. III. To add terms which ar e n o t all like terms.Add togeth er the terms wh ich a r e like terms by the r ul ei n th e seco nd case

, a nd p ut down th e other terms eachp r eceded by i ts p r op er sign .

For example ; add together4a + 5 b—7c+ 3d

, 3a 9a — 2b—c—d,and —a + 3b+ 4 e - 3d + e.

It is conven ien t to arrange the terms in column s,so

]tha

tlike terms shall stand in the same column ; thus we

1avo

AD D ITION :‘

1 7

4a + 5b—7c + 3d

3 a b+ 2c + 5d

9a — 2b c d

— a + 3b+ 4c - 3d + e

I5 a + 5 b— 20 + 4d+ 6

Here the terms 4 a,3a

,9a

,an d —a ar e all like terms ;

the sum o f the positive coefficien ts IS 1 6 , there 1s o n e termwith a n egative coefficien t, n amely — a , o f which the co

efficien t is 1 . The differen ce of 1 6 an d 1 is 1 5 ; so thatwe obtain + 1 5 a fi om these like terms ; the sign mayhowever be emitted by Art . 3 8 . Similarly we have5 b—b An d so on .

4 7. In the foll owing examples the terms are arrangedsuitably in column s :

x3+ 2x2 3x + 1 a

2+ ab+ bz—c

4x3 + 7x2+ x — 9 3 a

"’ —3ab— 7b2

2x3 + x‘z 9x + 8 4 a

“- l- 5 ab+ 9b2

3x" x2 + 10x — 1 a

2—3ab—3b2

x — l

In the first example we have in the fir st columnx9+ 4x3 — 2x3 —3x3 , that is 5 x 3 —5 x 3, that is, n othin g ; this

is usually expressed by saying th e term s wh ich in e o lr e x“

can cel each other .

Similarly, m the second example, the terms which involve ab can cel each other ; and so also do the terms whichinvolve b“

7x’ — 3xy to

3x” y2+ 3x

—2x2+ 4xy + 5 22

x —‘Zy

—7xy g’+ 9x — 5 y

+ 4y2 2x

12x2 Gr y 7p2 l 8g

18

a

g

o:

12.

13 .

EXAMPLES. V

EXAMPLES. V.

Add togeth er

3a—2b, 4a—5 b, 7a— l l b, a + 9b

4x2— 3y2,2x“ 5 51 wh y

”,

5 a + 3b+ e, 3a + 3b+ 3c, a + 3b+ 5 c.

3x + 2y—z

,2x—2y + 2z,

7a — 4 b+ c, Ga + 3b—5 c,- 12a + 4c.

x—4 a + b, 3x + 2b, a — x—5 b.

a + b— c,b+ c— a

,c + d — b

,a +b—e.

a + 2b+ 3c, 2a—b—2c, b—a — c,

c—a—b.

a — 2b+ se— 4d, 3b—4c + 5 d— 2a, 5 c—6d+ 3a—4b,

7d—4a + 5 b —4 c.

—4x2 + 5 x — 3 , 2x3—7x

2—14x 4—5 ,

x4—2x3 + 3x2, x

3+ x

2+ x , 4x4 + 5x3

,2x2+ 3x—4 ,

—3x2— 2x — 5 .

3a9b 3ab2 b3, 2053 5 a

’b 6abz 7b“,

a3 —ab2+ 2b

3.

x3—2ax2 - l—a2x -l- a

3,

x3+ 3ax

’,2a 3 — az

g— 2x3.

2ab Bax22d

‘lx,12ab max

”6a

2x,

—8ab+ ax3— 5 e2x .

x2+ y

4+ z ,

4x2—5z3, s.i

3 —7y4+ 10z3 , 6y

4—6i 3.

3x’ —7,—5 y + 8,

10xy + 8y2+ 9y , 5 x9—6xy + 3y

2+ 7x —7y + 1 1 .

x4 — 4x 3y

- i 6x1‘

3/2— 4xJ + 1J

4,4 .x

3y—l 2 i’ y

5’4- 12xy

3—4y‘,6x

’

y2—12xy

3+ 6y

4,4x3/

3—4g/4, y

‘.

—x‘-

’

J xyx—x i’

z,

— xJ’"—J’z—xJz,

xix - ry

zz -i- fi —xyz

-

y zh xs".

SUB TEAo rzozv. 19

VI. Subtraction .

48. Suppose we have to take 7+ 3 from 12 ; the resultls the same as if we first take 7 from 12, and then take 3from the remainder ; that is, the result is den oted by12 —7— 3 .

Thus 12 —7 - 3 .

Here we enclose 7 3 in brackets in the first expression ,because we are to take the wh ole of 7+ 3 from 12 ; see

Art. 29.

Similarly 20 — 5 —4—2.

In like mann er, suppose we have to take b+ c from a

the result is the same as if we first take b from a, and

then take c from the remainder ; that is, the resul t isden oted by a b—c.

Thus a —b—e.

Here we en close b c in brackets in the first expression ,because we are to take the whole o f b+ c from a .

S imilarly a —b—c—d.

4 9. Next suppose we have to take 7— 3 from 12. Ifwe take 7 from 12 we obtain 12—7 ; but we have thustaken too much from 12

,for we h ad to take, n o t 7, but 7

dimin ished by 3 . Hen ce we must in crease the result by 3 ;and thus we obtain 12 — 7 + 3 .

Similarly 12 —7

In like mann er,suppose we have to take b—c from a .

If we take b from a we obtain a —b; but we have thustaken too much from a

,for we h ad to take, n ot b, but 6

dimin ished by 0. Hen ce we must in crease the result by c ;and thus we obtain a (b —c)= a b c.

Similarly a—(b+ c—d)= a—b—c+d.

5 0. Con sider the examplea — (b+ c—d)= a—b—c+d ;

that is , if b c d be subtracted from a the result is

20 TBACTI ON.

a —b Here we see that, in the expression to besubtracted there is a term —d , and in the result there isthe corresponding term + d ; also in the expression to besubtracted there is a term + c, and in the resul t there is aterm — c also in the expression to be subtracted there is aterm b, and in the resul t there is a term b.

From con sidering this example, and the others in thetwo preceding Articles we obtain the following rule forSubtraction : ch ange th e s ign s of a ll th e terms in th e ex

p r ession to be subtr acted, and th en collect the terms as inAddition .

Fo r example ; from 4x —3g+ 2z subtract 3x y +z.

Change the signs o f all the terms to be subtracted ; thuswe obtain —3x + y z ; then collect as in addition ; thus

4x — 3y + 2z— 3x +g

—z= x—2y +z.

From 3x4 + 5x3 — 6x2 7x + 5 take 2x4 2x3 + 5 x2—6x—7.

Change the signs o f all the terms to be subtractedand proceed as in addition ; thus we have

3x4 + 5 x3 6x2— 7x + 5

—2x4 + 2x 3 5 x2+ 6x + 7

x4+ 7x3— l 1x 2 x + 12

Th e beginn er will find it pruden t at first to go throughthe operation as fully as we have don e here ; but he maygradually accustom himself to putting down the resultwithout actually changing all the signs, but merely supposing it done

5 1. We have seen thata—(b—c)= a—b+ c.

Thus corresponding to the term —c in the expressionto be subtracted we have + 0 in the result. Hence it isn o t un common to find such an example as the followingproposed for exercise : from a subtract —c ; and the resultrequired is a + c. Th e beginn er may explain this in themann er o f Art . 4 1

,by con siderin it as having a mean ing,

n o t 111 itself, but in conn exion Witti some other parts o f an

algebraical operation .

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22 BRACKE TS.

VII. B r ackets.

5 2. On accoun t of the exten sive us e which is made o f

brackets m Algebra, it IS n ecessary that the studen t shouldobserve very carefully the rules respecting them,

and weshall state them here distin ctly.

PVh en an exp r ession with in a p air of brackets is p i eceded by th e s ign the br ackets may be r emoved.

Wh en a n exp r ession with in a p air of br ackets i s p r eceded bg th e s ign th e br ackets m ay be r emoved if thesign of ever y term with in th e br ackets be ch anged.

Thus, for exampla — b+ (c —b+ c—d+ e,a — b— (c —b—c + d - e.

Th e second r ule h as already been illustrated in Ar t . 5 0

it is in fact the r ule f o r Subtr action . Th e first rule mightbe illustrated in a similar mann er.

5 3 . In particular the studen t must n otice such statemen ts as the following :

—d) : —d,

These must be assumed as rules by the studen t, whichhe may to some exten t explain , as in Art . 4 1 .

5 4 . Expression s may occur with more than o n e pair ofbrackets : these bi ackets may be removed in succes sion bythe preceding rules begin n ing with th e in side p a ir . Thus,f0 1 example

,

—d,

a + {b—(c —c + d,

a — {b+ c— d}= a — b—c+ d,

a — {b— (c —{b — b+ c—d.

Similarly,a—[b— {c—(d — [b—{c

a—[b—c+ d — e]= a—b+ c—d + e.

It Will be seen in these examples that, to preven t confusion between various pairs o f brackets, we use brackets

BRAOKE TS . 23

o f differen t shap es ; we might distinguish by using bracketso f the same shape but of differen t sizes.

A vincul um is equivalen t to a bracket ; see Art. 30.Thus

,for example

,

a - [b —(d—e —[b —(da—[b—{c—d + e —[b- c+ d—e+f ]a ~ b+ c — d+ e—f .

5 5 . Th e beginn er is recommended always to removebrackets in the order shewn in the preceding Article ;namely, by removing first the inn ermost pair

,n ext the in

n e rmo st pair o f all which remain,and so o n . We may h ow

ever vary the order ; but if we remove a pair of bracketsin cluding an other bracketed expression within it, we mustmake n o ch ange in th e sign s of th e in cluded exp r ession .

In fact such an in cluded expression coun ts as a single term.

Thus,for example

,

—d,

a + {b— (c —(c — c+ d,

a —b— (c—d)= a —b—c + d,

a —{b— (c —b+ (c—d)= a—b+ c— d.

a —[b— {c—(d a—b+ {c— (d — e)}= a —

'

b+ c— (d— e)= u—b+ c—d+ e.

And in like mann er, a —[b—{c— (d— e

a—b+ {c— (d—e — b+ c— (d— e—f )

a—b+ c —b+ e—d + e<fl

5 6 . It is often conven ient to put two o r more termsWithin brackets ; the rules fo r in troduc ing bracketsimmediately from those for removing brackets.Any n umber of terms in an exp r ession may be p u t

with in a p air of br ackets and th e sign p laced bef or eth e wh o le.

A n y n umber of terms in an exp r ession may be p ut

with in a p a ir of br ackets and th e sign p laced bef o r eth e wh ole

, p r ovided th e sign of ever y term with in th e

brackets be ch anged.

EXAMPLES. VII .

Thus for example,a—b+ c—d + e

a —b+ (c a

o r = a—(b—c+ d— e), or a — b — c+ d — e).

In like mann er more than o n e pair o f brackets maybe in troduced. Thus, fo r example,

a —b+ c - d + e=a— {b—c + d—e}= a —{b- (c

EXAMPLES. VII.

Simplify the following expression s by removing thebrackets and collecting like terms :

1 . 3a— b— (2a—b).0v 2. a—b+ c— (a—b—g)1 1Q

3 . 1 —a + a2) —a + a

2

4 . a + b+ (7a —b)—(2a—3b)5 . a — b+ c— (b —a + b)—(a

6 . 2x — 3y—3z—(x —(z—x—y )- 3 xt3:

7. a—{b—c— (d a £ 4 t (“ 6 - z

s. 2a—(2b—d)—{a — b—(2c fi t “L

9 . a—{2b—(3c+ 2b— a)}J L 10. 2a—{b—(a—2b)}1 1 . 3a —b) — (a —L L,

12. 7a — [3a—{4a—(5 a S' ”

L

13 . 3a— [b14 . 6a—[4b {4 a—(6a l / 1

1 5 . 2a — [5 b —6b) + 5 c—{2a - n.l t1 6. a — 3 a a r e,

17. 16 —2x

18. 1 5x — 5 x—(se

19. 2a—[2a — {2a— (2a -i 2ai

20. 16—x—[7x—{sx - (9xr?“

21 . 2x—[3y—{4x—(5 y — 6x22. 2a —c)— 4c + {2a—(3b—c E;

23 . a — [5b —{a - (5 c—2c —b— 4b) + 2a—(a —2b—4

-

m 1“L

24 . x4

[4xl {6x2 (xi 4x3 6x

3 4x b y

M ULTIPLICATI ON. 25

VIII. M ultip lication .

5 7. The studen t is supposed to kn ow that the producto f any number o f factors is the same in whatever order thefactors may be taken ; thus 2 x 3 x x 5 x x 5 x 2 ;and so o n . In like mann er abc= acb= bca

,and so 011 .

Thus also c(a + b) and (a + b)c ar e equal , for each den otes the product o f the same two factors ; on e factorbeing c

,and the other factor a b.

It is conven ien t to make three cases' in Multiplication,

namely, I. Th e multiplication o f simple expression s ; II. Themul tiplication o f a compoun d expression by a simple express1on ; III. The multiplication o f compound expression s. We shall take these three cases in order.

5 8. I. Suppose we have to multiply 3a by 4b. Th eproduct may be written at full thus 3 x a x 4 x b, or thus3 x 4 x a x b; and it is therefore equal to l 2ab. Hence wehave the following rule for the mul tiplication of simple ex

pression s ; m ultip ly togeth er th e n umer ica l coefi cien ts

and p ut th e letter s af ter th is p r oduct.

Thus for example,7a x 3be= 2l abe,

4a x 5 bx 3c= 60abc.

5 9. The p ower s of the same n umber ar e mul tip l ied

togeth er by adding th e exp on en ts.

Fo r example,suppose we have to multiply a3 by CL2.

By Art . 16,

and a2= a x a ;

therefore a3 x a

'

x a x a x a x

Similarly,In like manner the rule may be seen to be true in any

other ca

se.

26 M ULTIPLIC’A TIOM

60. II. Suppose we have to multiply a +b by 3 . We

have

Similarly, 7(a b) 7a 7b.

In the same mann er suppose we have to mul tiply a +bby e. We have

c(a b)= cd cb.

In the same mann er we have3 (a — b)= 3a—3b, 7(a b) 7a—7b, c(a b)= ca—cb.

Thus we have the following rule for the mul tiplicationo f a compound expression by a simple expression ; m ultip lyeach term of th e comp ound exp r ession by th e s imp le ex

p r ession ,an d p ut th e sign of th e term bef or e th e r esult ;

a nd collect th ese r esults to f o rm th e comp lete p r oduct.

6 1 . III. Suppose we have to mul tiply a +b by c+ d.

As in the second case we have(a + b)

also

th erefore (a b)(c d) ac ad+ be bd .

Again ; mul tiply a —b by c d.

(aalsotherefore(a (bc+ bd) = ac+ ad—bc bd.

Similarly ; multiply a b by c d.

(a + b)(c—d) : (c —d(a + b)ca + cb cb—da—db.

Lastly ; multiply a—b by c —d .

(a—b)(c—d) : (c— d)a — (c—d)balso (c—d)a =ac —ad

, (c—d) b: bc—bd

therefore(a —b)(c—d)= ac—ad (be—bd)= c c—ad—bc bd.

Let us n ow consider the last result. 13y Art. 3 3may write it thus,

(+ a —d) = + ac—ad bc +bd.

M ULTIPLI0A TION . 27

We see that corresponding to the + a which occursin the mul tiplicand and the c which occurs in the mul tiplier there is a term ac in the product ; corresponding tothe terms + a and —d there is a term - ad in the product ;corresponding to the terms —b and + c there is a term—bc in the product ; and corresponding to the terms —band —d there is a term bd in the product.Similar observation s may be made respecting the other

three results ; and these observation s are briefly collectedin the following importan t rule in multiplication : l ike s ign sp r oduce an d un l ike sign s This rule is call ed theR ule of S ign s, and we shall often refer to it by this n ame.

62. We can n ow give the gen eral rule for multiplyingalgebraical expression s ; mu ltip ly each term of th e multi

p lican d by each term of th e mu l tip lier ; if th e terms h ave

the same sign p r efix th e sign to th e p r oduct, if th eyha ve difl

’

er en t sign s p r efix th e sign th en col lect th ese

r esults to f orm th e qomp lete p r oduct.

For example ; multiply 2a + 3b— 4 c by 3a—4b. Here(2a +Sb—4c) (3a — 4b)= 3a (2a + 3b— 4 c)—4b (2a + 3b

—4c)= 6a2 9ab 12ac (8ab 12b2 1 6bc)= 6a

2+ 9ab 1 2ac— 8ab 1 2b2+ 1 6be.

This is the result which the rule will give ; we maysimplify the result and reduce it to

6a2+ ab 12ac 12b7 + 16bc.

We might illustrate the rule by using it to multiply6—3 2 by 7 3 — 4 ; it will be found that o n w orking bythe rule, and collecting the terms, the resul t is 30, that is5 x 6, as it should be .

6 3 . Th e studen t will sometimes find such examples as

the following proposed : multiply 2a by—4b, or multiply

- 4c by 3a, o r multiply —4 c by — 4b.

The results which are required are the following,2a x —4b= 8ab

,

- 4 c >< 3a = — l 2ac,—4 c x —4b 1 6bc.

28 M ULTIPLICATION .

Th e studen t may attach a mean ing to these Operationsin the mann er we have already explain ed ; see A rticle 4 1Thus the statemen t — 4c ><—4b= 1 6bc may be unde1

stood to mean , that if —4c occur among the te rms o f amul tiplican d an d —4b occur among the terms o f a multiplier

,there will be a term 16bc in the product correspond

ing to them.

Particular cases o f these examples are2d x —4 : — 4 —8, 2 x —1 — 2.

64 . Sin ce then such examples may be given as thosein the precedingA 1 ticle , it becomes n ecessar y to take ac

coun t o f them in o ur rules ; and accordingly the rules formultiplication may be conven ien tly presen ted thusTo multiply simple terms ; mu ltip ly togeth er th e n u

mer ical cbefi cie1i ts, p ut th e letter s af ter th is p 1 oduct an d

determin e th e sign by th e Rul e of S igns.

To multiply expression s ; mu ltip ly each ter m in o n e

exp r ession by each term in th e oth er by the r ule f o r mul

tip ly ing simp le terms,an d collect th esep ar tial p r oducts to

f orm th e comp lete p r oduc t.65 . We shall n ow give some examples ofmul tiplication

arranged in a convemen t form

a2

ab

ab b“

ae+ 2ab+b2 x

3 + 2x2 —3 .v

a3 4 a

Qb ab2 3a“ 4a 3b 5 1t2

crib—ab” b3 6a

3b 8a?b2l oubi

9a‘-

’

b’

12ab3 1 5 6‘

3 a‘10a3b 22a3b

‘

3 220 b3 1 5 64

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30 Ill ULTIBLICA TION.

x4—3x’

+ 2x + 1

x3— 2x —2

x7—3x5 -1- 2x4 - l- x

3

—2x 5 6x3 — 4x2—2x

— 2x4 + 6x2 — 4x—2

x7— 5 x

5+ 7x3 + 2x

2— 6x — 2

Mul tiply a2+ b2+ c‘3—ab - bc— ca by a + b+ c.

A rrange according to descending powers of a .

a2—ab—ue +b2—bc +e

a b c

as

agb a

2c ab” abc ac

2

atz abc 63 620+bc’

abc a o2 b’c—bc’ + c3

3 abc b3

This example might also be worked with th e aid ofbrackets

,thus,a2

a (b+ c)

a3 + a (b

2—bc + c?)—bc + c

'

Then we have a(b" bc —a (b c) (b c)

a {bi—bc + c2 (b

a{bz— bc+ c

2—(b'

—a{b

”—bc+ cg —2bc —3abc ;

T hus, as before, the result is a3 b3 03 - 3abo.

III ULTIPLIC’A TION . 3 1

Multiply together x a,x b, x c.

xz— e x

—bx + ab

c

(a b)x2

abx

cx2(a b)ex abc

—abc

The studen t should n otice that he can make two exercises in multipl i cation from every example in which themultiplicand and multiplier ar e differen t compound ex

pression s,by changing the origin al multiplier in t o the

multiplicand,an d the orig1nal multiplicand 1n to multiplier.

Th e result obtain ed should be the same, which will be a

test o f the correctn ess o f his work.

EXAMPLES. VIII.

Multiply1 . by 4x

”. 2. 3a4 by 4a

5. 3 . 2a2b by 2ab

’

4 . 3x 3y2z by 5 x

4y3z2. 5 . 7x"y

zby

6 . 4 a2 3b by 3ab. 7. 8a2—9ab by 3a2.

8 3x2 4g

2 5x2 by 2x?y .

x“

y3y3z4

z4x2 by x

g

ygz’.

2xy2z3 3x2y

3z 5 x3yx

2 by 2xyzz.

2x y by 2y x .

2x3 + 4x2+ 8x + 16 by 3x

—6.

x3+x

2+x—1 by x

—l .

EXAMPLES . VII I.

1 -1- 4x - ' 10x2 by 1

1 5 . x3 — 4x

’+ 1 1x — 24 by x

2+ 4x + 5 .

1 6 . x3+ 4x

2+ 5 x ~ 24 by x

9—4x + 1 1 .

17. x3 —7x

2+ 5 x + 1 by 2x

2—4x + 1 .

18 . x3+ 6x

2+ 24x + 60 by x

3 —6x‘

1 2x + 12.

19. x3 —2x

‘3+ 3x — 4 by 4 x 3 + 3x

2+ 2x + 1 .

20. x4—2x 3 + 3x2— 2x + l by

2 1 . x2— 3ax by x + 3a .

22. a‘3+ 2ax — x

2 by a2+ 2ax + x

2

23 . 2b‘3+ 3ab— a

2 by 7a— 5 b.

24 . a2—ab+ b2 bv a

g+ ab— b2.

25 . a2— ab+ 2b2 by

26. 4x2 — 3xy—y2 by 3x

— 2y .

27. —x4y + xy

4—y5 by x + y .

28 2x2+ 3xy + 4y9 by

x9+ y

2—xy + x + y— 1 by x + y

—1 .

x4 2x3y 4x “

°

y’Sxy

3 my by x 2g.

8 1x‘+ 27x3y + 9x

9y2+ 3xy

3+ y

‘ by 3x—y .

x + 2y—3z by x

— 2y 4- 3 r

a2

- ax + bx + b2 by a + b+ x .

a2+ b2+ c

’ — bc—ca —ab by a + b+ c.

3 5 . by a2

3 6. a2—2ab+ b2+ c2 by a

2+ 2ab+ b

’—c’.

Multiply the following expressions together

x + a a’s- a

"

x + c

— ax + a’

x‘

x + 2m

D I V[ S] 01V. 3 3

IX. D ivision .

68. D ivision , as in Arithmetic, 1s ‘

tlie inverse of Mul tiplication. In Multiplication we determ in e the productaris ing from two given factors ; in D ivision we have giventhe product and o n e o f the factors, and we have to determin e the other factor. The factor to be determin ed iscalled the quotien t.

The presen t section therefore is closely conn ected withthe preceding secti on

,as we have n ow in fact to undo the

operation s there performed . It is conven ien t to makethree cases in D ivision

,n amely, I. The division of on e

simple expression by an other ; II. The division o f a com

pound expression by a simple expression ; III. The divisionof on e compound expression by another.

69. I. We have already shewn in Art . 10 how toden ote that o n e expression is to be divided by an other.Fo r example, if 5 a is to be divided by 20 the quotien t isindicated thus : 5 a + 2c, o r more usually2

5

5.

It may happen that some o f the factors o f the divisoroccur in the dividen d ; in this case the expression fo r. thequotien t can be simplified by a prin ciple already us ed inA rithmetic. Suppose, for example, that 1 5 a2b is to bedivided by 6bc ; then the quotien t is den oted by

1

2306

.

Here the dividend 1 5 a‘2bz 5 a

2x 3b; an d the divisor

6bc= 2c x 3 b ; thus the factor 3b occurs in both dividendand divisor. Then , as in Arithmetic

,we may r emo

ge

this common factor,and den ote the quotien t by

5

2

6

;

3 4 .D I VISION.

‘

It may happen that all the factors which occur in thedivisor may be removed in this man n er. Thus suppose

,for

example, that 24abx is to be divided by 8ax :24abx 3b x 8ax

Sax

70. Th e rule with respect to the sign o f the quotien tmay be obtain ed from an examin ation o f the cases whichoccur in Multiplication .

For example, we have4 ah x 3c= 12abc °

therefore 4ab.

4abx —3c= l 2abc °

therefore

4abx 3 c 12abc ;

therefore

- 4abx 3c= 120bc

therefore I2abc4ab.

Thus it will be seen that the R ule of S igns holds inD ivision as well as in Mul tiplication .

71 . Hen ce we have the following rule for dividing o ne

simple expression by an other : Wr ite the d ivide nd overthe divisor with a l in e between th em ; if th e exp r ession sh ave common factor s, r emove th e common f actor s ; p r efixth e s ign if th e exp r ession s ha ve th e same sign and the

sign if th ey h ave d iffer en t s ign s.

72. O n e p ower of any n umber is divided by ano th er

p ower of th e same n umber , by subtr acting the index Qfth e la tter p ower f r om th e in dex of the f ormer .

D I VISION 35

For example, suppose we have to divide as by a3

By Art . 16 , a5z a x a x a x a x a

,

a3 = a x a x a ;

5a x a x a x a x atherefore = a x a= a

2=a x ( t x a

O X C X G X C X C X G X G

C X C X C X OSimilarly

64

— C X C X C= 03 = C

In like mann er the rule may be shewn to be true in anyother case .

O r we may shew the truth o f the rule thusby Art . 5 9, c

4x c

3 = c7,

therefore

73 . If any power of a. number o ccurs in the dividendand a higher power o f the same number in the ; divisor, thequotien t can be simplified by Arts. 71 , and 72. Suppose,for example, that 4 abzis to be divided by 3 cb5 ; then the

2

quotien t is den oted by 36

0

1

55 Th e factor b2 occurs in bothdividend and divisor ; this may be removed, and the quo

4a 4 abz 4atl ent den oted by

3 cb3 3 cb5 3c

74 . II. Th e rule for“ dividing a compound expressionby a simple expression will be obtained from an examinah ea d the corresponding ease in Mul tiplication .

For example, we have

(a b) c= ac—bc ;

ac—bc_

c

(a —c= - ac + bc ;

therefore a b.

therefore

3 6 D IVISION.

Hence we have the following rule for div iding a‘

. compound expression by a simple expression . divide each term

of th e d ividen d by the divisor,by the r ule in the ji1 st

case, and collect th e r esults to f orm the comp lete quo tien t.4a3 3abc a

?o

aFo r example, 4a2 3bc ac.

75 . III. To divide on e compound expression byan other we must proceed as in the Operation called LongD ivision in Arithmetic. The foll owing rule may be given .

A r r ange both dividend an d divisor acco r ding to ascend

ing p ower s of som e common letter , o r both accor ding todescen d ing p ower s of some comm on letter . D ivide th e

fir st term of th e dividen d by th e fir st term of the divisor ,an d p ut th e r esul t fo r th efi r st term of the quotien t ; m ul

l ip ly th e wh ole divisor by th is term an d subtr ac t the

p r oduct f r om th e dividen d. To th e r ema in der join as

man y terms of th e dividen d,taken in order , as may be

r equir ed, an d r ep eat th e wh o le op er ation . Con tin ue th ep r ocess un til a ll th e terms of th e d ividen d ha ve been

taken down .

The reason fo r this rule is the same as that for therule Of Long D ivision in Arithmetic , nahi e ly, that We '

maybreak the dividend up in to parts and find h ow Often thedivisor is con tain ed in each part

,and then the aggregate

o f these results 18 the complete quotien t.

76 . We shall n ow give some examples Of D ivisionarranged in a conven ien t fO 1m.

—bi (a- b

a2+ ab a

’+ ab

a—b) a2

- b2 (a + b— 3x (x

—l

az— ab (o i

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38 D IVISION .

pressed in ways similar to those used in Arithmetic ; thuswe may say that

a‘

- l- 2b2

a + b=a +b+

that is,there is a quotien t a +b, and a fractional part

In gen eral, let A and B den ote two express ions , andsuppose that when A is divided by B the quotien t is Q, andt h e remainder R ; then this resul t is expr essed algebrai

cally in the following ways,A q +R

,o r A—n R ,

The studen t will observe that each letter here may r epresen t an expression

,simple o r compound ; it is often

conven ien t for distinctn ess and brevity thus to represen tan expression by a single letter.

We shall however con sider algebraical fraction s in subsequen t Chapters, and at presen t shall confin e ourselves toexamples o f D ivis ion in which the operation can be exactlyperformed.

78. We give some more examples :D ivide x7—5 .v

5+ 7.e 3 + 2x

2—6.v — 2 by 1 + 2v

Arrange both dividend and divisor according to descending powers o f x .

-i- 1) .v7

- 5 x5 —6.v—2ka3 —2.v—2

'

w7 a“

2.v5 — 2.v

4+ 2x’

2x 5 6x3 2x

+ 6 .v’ — 4 .v— 2

+ 6a3 —4x—2

D I VISION. 39

D ivide a3 +b3 + 03—3abc by a +b+ c.

A rrange th e dividend according to descending powerso f a .

a + b+ c) a3 —3abc+ b3 + c3 ta

2—ab—ac+ bz

a3+ a

gb+ a2c

—d"’c ab" 2abc”c abc— ac

2

ab2 abc+ acz+ lrs

abz b3 bzc

It be seen that we arrange these terms accordingto descen din g powers o f a ; then when there ar e twoterms, such as azb and a

2c, which in volve the same power o f

a,we select a n ew letter, as b, an d put the term which

con tain s b before the term which does n o t ; and again , o f

the terms abzand abc, we put the former first as in volvingthe higher power o f b.

This example might also be worked, with the aid o f

brackets, thus :

a + b+ c) a3

- 3 abc+ b3 + c3

La2 —be + e'

—3 abc + b3 + c3

a(bz

a(bz bc

40 EXAMPLES . JX .

D ivide a‘3 —abc by x—c.

x c) .v3

ac r z

art — ex

”

- abc

(a b).v2

(a b)e.v

abm

Every example o f Multiplication ,in which th e multi

plier and the multiplicand ar e differen t expressions, willfurn ish two exercises in D ivision ; because if the productbe divided by either factor the quotien t shoul d be the otherfactor. Thus from the examples given in the secti on o n

Multiplication the studen t can derive exercises in D ivision ,and test the accuracy o f h is work. And f1 om any exampleo f D ivision , in which the quotien t and the divisor ar e

differen t expression s, a secon d exercise may be obtain edby making the quotien t a divisor o f the div idend, so t hatth e n ew quotien t ought to be the original divisor.

EXAMPLES. IX.

D ivide1 . 1 5 a“5 by 2 . 24a6 by 8e 3 . 3 . 18x33}

? by fizzy .

4 . 24a4b5c6 by 5 . 20a 4b4d 3y3 by 5 b

‘3a3y .

6 . 4 .v3 — 8.v

2+ by 7. 3a4 — 1 2a3 + 15a ’ by

8 . (fig 3 .v"’

y24 am

" by my .

9 . l 5 a 3b3 3a‘3b

‘2 12ab by Bab.

l 0. 60a3b3c2 4 Sa‘3b4c2 3 6a

2b2c‘ QOabc6 by 4abc’.

l l . x2—7x + 12 by m

—3 . 12.

—72 by x + 9.

1 3 . 2x 3 —9 by 2x—3 .

1 4 . 6a3 + l 4 .v°—4 .e + 24 by 2.v+ 6 .

1 5 .—l by

— 1 .

1 6 . 7.v"

5 833—2 1 by 7.v— 3 .

EXAMPLES . IX . 4 ]

x6— 1 by .v — 1 . 18 . a3—2ab2+ b3 by a

—b.

x4 —8 1y

4 by x- 3y .

x4 —

.vy3 by x

—y .

x5

—y5 by x

—g. 22. a

5- 1- 32b5 by a + 2b.

2a4 + 27ab3 — 81b4 by a + 3b.

by x3+ y

3.

x5

xiii/3 Qxy

" 3y5 by x

3—y3

a'4 —12.v + 6 by x

2—3x - l- 3 .

a4+ a3 —9.v

2— 16 .v — 4 by afl + 4x + 4 .

x4 — 1 3f - l- 3 6 by x

2+ 5 x + 6 .

m4 + 6 4 by fi + 4x + 8 .

bya4 +w3—24x2—3 5x + 5 7 by a

“+ 2.v —3 .

1 — ar — 3a" — .z'5 by

J's

- i- 1 by w”

a4+ 2a

9b2 + 964 by a2—2ab 3b?

a“—b6 by a

3

a‘+ 2x

5- 4aA—2a3 4 - 12x2—2x—1 by x2+ 2w - l .

by w4 — 2x3 —i- 3x

2—2a'4- 1 .

mm - l- a's— Qby w

4+ a

2+ 1 .

{IF— (a —abc

by az

a' —b

‘3) l

ei - 6

2 by dx2—bx—l- c.

x4—a3y —wy

3-i- y

“by x2+ xg + y

2

org— 3xy

—y3 — 1 by w

—y—l .

4 9x2+ 21xy + 12yz— l 6ag by 7x + 3y

— 4z.

a2+ 2ab+ b

'2— c2 by a + b

— c.

a3 + 8b3 + c3 —6abc by a

2+ 4b2+ c

?—ac—2ab - 2bc.

by a + b+ c.

— b) + abc by a +b+ c.

.’c3 —( fife — (db+ ab2 by x

— a -i- b.

(.v - l by x + y—z.

by

GENERAL RESULTS

X. Gen er al Results in M ultip lication.

79 . There are some examples in Multiplication whichoccur so often in algebraical operations that they deserveespecial n otice.Th e foll owi ng three examples are o f great importan ce.

a + b

a —b

a22ab a

2 2ab b?

Th e first example gives the value of (a b) (a b), thatis, of (a thus we have

(a + 2ab b”.

Thus th e squa r e of the sum of two n umber s is equa l toth e sum of th e squar es of th e two n umber s in cr eased bytwice th eir p r oduct.

Again , the second example gives

(a —b)2= a2—2ab+ bz.

Thus th e squar e of th e difi'

er en ce of two n umber s a‘

s

equal to th e sum of th e squa r es of th e two n umber sdimin ish ed by twice th e ir p r oduct.Th e last example gives

(a + b) (a — b)= a2— b

Thus th e p r oduct of the sum a nd def er ence Qf twon umber s is equal to th e dzf er en ce of their squa r es.

80. The results of the preceding Article furn ish a

simple example o f o ne of the uses o f Alg ebra ; we maysay that Algebra enables us to p r ove gen e ral theor ems

r esp ecting n umber s, and also to exp r ess tho se theo r emsbr iefly.

IN M ULTIPLI0A TI 01V. 4 3

Fo r example,the result —b)= a

z—b2 is provedto be true, and is expressed thus by symbols more compactly than by words.

A gen eral result thus expressed by symbols is oftencalled a f o rmula.

We may here indicate the mean ing of the sign i

Wh l ch 13 made by combin ing the sign s and and whichis called the double sign .

S in ce (a b)2= a

z 2ab b2,and (a b)

2= a2 2ab b2,

we may express these results in on e formula thus :(a i b)

2= a2 i 2ab bz,

where i indicates that we may take either the sign orthe sign keep in g th r ough ou t th e upp er sign o r thelower sign . a i b is read thus, a p lus o r m in us b.

”

82. We shall devote some A rticles to explain ing theuse that can be made o f the formulae o f Art. 79 . We shallrepeat these formul ae

,and n umber th em f o r th e sake of

easy and distin ct r ef er en ce to th em .

(1 )(a — b)2

-

a,2

2ab+ b2

(2)- b) = a

2— b2 (3 )

8 3 . Th e formulae will sometimes be o f use in Arithmetical calculation s. For example ; required the differenceo f the squares of 127 and 123 . By the formula (3 )

(123V: (127 25 0 x 4 1000.

Thus the required number is obtain ed more easily thanit would be by squaring 127 and 123, and subtracting thesecond result from the first.Again , by the formula (2)

—60+ 1 = 84 1 ;and thus the square of 29 is found more easily than bymultiplying 29 by 29 directly .

O r suppose we have to multiply 5 3 by 4 7.By the formula (3 )5 3 x

44 GENERAL RES ULTS

84 . Suppose that we require the square o f 3x + 2y.

We can o f course obtain it in the ordin ary wa that is bymul tiplying 3x + 2g by 2g. But we can ahain an other way, n amely, by employing the formula.The formula is true whatever number a may be , and whate ver number b may be ; so we may put 3x for a

,and 231

fo r b. Thus we obtain

(3x 2g)”(3x )

2 2 (3a: 2g) (2g)2 9x

2 12xy 4g”.

Th e beginn er will probably think that in such a case hedoes n o t gain any thing by the use o f the formula, fo i'he will believe that he could have obtain ed the requiredresul t at leas t as easily

'

and as safely by common workas by the use o f the formula. This n otion may be correctin this case, but it will be found that in more complexcases the formula will be o f great service.

85 . Suppose we require the square o f D en ote .v + y by a .

Then and by the us e o f (1) we have

ThusSuppose we require the square o f p q r s . D en oteg by a and r — s by b ; then p —

q+ r

By the use o f ( l ) we have—q) (r

Then by the use of (2) we express (p —q)

’and (r

Thus (p —a r — s)2

—qr + qs) 4—r i — 2r s ~i- s2

— 2p q- 2p s

—2qr—2r s.

Suppose we require the product o f p—q + r

—s and

p—q—r + a

Let p—q z a and r —s= b ; then

p—q + r snd p

—g—r 4—s z a —b.

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4G RES ULTS IN M’

ULTIPLICA TION .

(a2 2ab b

'3 ~ cz) (c

ea2 2ab b?)

{2ab (a2 b2 c'

{2ab (a2 b2

(24 5? (a2 5 2 59

2

{(a2 2 (a

2 b?) 62

c‘}

eaib2 (a2 b?)

2 2 (a2 b?) c

2c“

4a2b2 a“ b" 2b2c2 c

‘

2a2b2 2b2c2 2a‘zo2

a“ b4 C

4.

88 . There are other results in Multiplication which areof less importan ce than the three formulae given in A rt . 82 ,but which ar e deserving o f atten tion . We place them herein order that the studen t may be able to retei to themwhen they ar e wan ted ; they can be easily verified byactual multiplication .

(a + b) (ai

(a—b) —b3,

(a b)3: (a b) (a

? 2ab b?) as 3a2b 3ab2 b3

,

(a b)3 = (a —b) (dz—2ab b?) = a3 3 a

9b 3ab2 b3 ,

—a3 + b3 + c

3+ 3a2(b+ c) Gabe.

89 . Useful exe1 c1ses in Multiplication ar e formed byi equiiing the studen t to shew that two expression s agree ingiving the same 1 csul t. For example, shew that

(a—b) (b—c) (c—a) = a (c b) +b2 (a - a).

If we multiply a - b by b— c we obtain

ab b2 ac be

then by mul tiplying“this result by c a we obtain

cab cb2 ac2 be“1 a

2b abz age abc

,

that is a2 (c b) b’ (a c) c

2(b

Again ; shew that (a (b —c)3 + (c ayi

i

2 (C—b) (0—a) + 2 (b—a ) (b —b) (a

EXAMPLES . X. 4 7

By using formula 2) o f Art . 82 we obtain(a —a)

2

=a2—2ab4- b2 4- b2—2be + cz+ e2—2ac+ a2

—ab— ac—bc).

(c—b) (c —ca

(b - a ) (b—e) z b2—ba — bc + ac

,

(a —b) (a —ab

therefore (c - b) (c —a) (b —b) (a —e)= a

2+ bg+ ez—ab— ao—bc ;

therefore (a —a)’

= 2 (c—b) (c —a) (b—c) ; i- 2 (d—b) (a—c).

I

Ex’

amPLEs. X .

Apply th e formulae o f Art . 82 to the following sixteenexamples in multiplication :

M y )? 2. (7x

2 5 y2

)2

(x2

4 . (a:25 x

3x 6 . (a: 2y 3a)2

lx2

xv y”) (x

2 my

(x2 my e

”) (w

2xv a

il.

(562 we a

?

) (x2

x31

w a?) (50

2ivy y

”)

1) (a'3—2x2 -l- 3x

(x (a’+ 6x + 1 3 . (a +b)

z(a2

(2x 3y)2

(4x212xy 9y

2

).

(ax by) (ax by) (a’lv ’

(as: by)”(an

4 8 EXAMPLES . X.

Shew that the following results ar e true17. (a

2 b?) (c2 d”

) (ac bd)2

(ad

18.

19. (a — b) (b—c) (c—a ) : bc (c—b) + ca (a—c)+ ab (b—a ).

20 . (a—b)3+ b3 —a 3 = 3ab (b— a ).

21 . —a (b+ c—a )—b(a + c—b)— c (a + b— c)

(a’

ab b2)2

(a2—

.ab bg)9= 4ab(a

’+

(a + b+ c)3— a

3—'

b3—c3 = 3 (a + b) (b+ c)

(a + b

b b —b(a + + c

—(b+ c— a )3—(a — b+ c)3 - (a +b— c)

3

—24abc.

27. a +b—c 2+

’

a - b -i - a)2

k

(a + b)2+ 2 (d

z—b”) (a

(a —b) (b—c) (c—a).

(a —~b)(2a)

3

—b) + (a4a

3.

—a9

)+ — b’)

(a — b) + (b —c)(c— a ) (a

z—c) (.c—a) = (a—b) (b—c) (a—c).

(c+ d)f

{(dx balz(de bar} {(aw by)

2(w W }(a4 b*) (x

4

(cy ba)’l(az er )

z(bx ay)2 (as: by cs )3

(a2 b2 c?) (4c

2y

FAcr os s. 49

XI. Factor s.

In the preceding Chapter we have n oticed somegen eral results in Multiplication ; these results may also beregarded in conn exion with D ivision , because every ex

ample ih Multiplication furn ishes an example or examplesin D ivision . We shall n ow apply some o f these resul tsto find what expression s will divide a given expression , o r

in other words to r eso lve exp r ession s in to th eir factor s.

9 1 . For example, by the use of formula (3) of Art. 82we havea4 b?) (c

g 5 2) (a + b) (0as b8 (a

4 b4) (a4 b4) 2 (a

4 b“) (a2 b?) (a b) (a b) .

Hen ce we see that a s b8 is the product of the fourfactors a

“ b4, a2 b2

,a b

,and a b. Thus a

8 b8 is

divisible by any o f these factors, o r by the product of anytwo o f them, o r by the product of any three o f them.

Again ,(a2+ ab+b2) (a

g—ab b2+ ab) (a2+ b2—ab)

(a2 bi)

z(ab

2a“ 2a2b2 b4 a

2b2 a“

afib2 b4

Thus a“ b4 is th e product o f the two factors

a‘3+ ab+ b2 and a

Q— ab+ b3,and is therefore divisible by

either o f them.

Besides the resul ts which we have already given , weshall n ow place a few more before the studen t.

92. Th e following examples in division may be easilyverified.

4

50—5r

x3+x

2y + xy

2+ yi

and so on .

5 0 FACTORS .

x “

%

—x3 — x

”e + xy

2—e3.

w5 —x4y + x

3y2—x

‘

as.

and so on .

a}? my y

2,

= x4$331 fr y $51

3at

and so on .

The studen t can carry on these operation s as far as

h e ple ases,and he will thus ga in confiden ce in the truth of

the statemen ts which we shall n ow make,and which are

strictly demon strated in the higher parts o f larger workso n Algebra . The following ar e the statemen ts :x"

-

y" is divisible by x —

y if n be a n y whole n umber1c

" —y" is divisible by a: y if n be an y even whole number

x"

y" is divisible l>y

’

m+ y if n be any odd whole number.We might also put in to words a statemen t of the forms

o f the quotien t in the three cases ; but the studen t willmost readily leam these forms by looking at the aboveexamples and

,if n ecessary, carrying the operation s still

farther.We may add that .v

“

+ y"

is n ever divisible by a'+ y o r

x —y , when n is an even whole number.

93 . Th e studen t will be assisted in remembering theresults of the preceding Article by n oticing the simplest

FAOTORS . 5 1

case in each o f the four results, and referring other casesto it. Fo r example, suppose we wish to con sider whetherx7—y7 is divisible by fc —y o r by x + y ; the index 7 is an

odd whole number, and the simplest case of this kind ism— y , which is divisible by x —

y , but n o t by {c + y ; so wein fer that x

7 —y7 is divisible by x —

y an d n o t by zc + y .

Again ,take xB— y s ; the index S is an even whole n umber

,

and th e”

simplest case of this kind is x2—y2, which is

divisible Both by w—y and .v + y ; so we in fer that afi -

ys

is divisible both by w—y and a y .

94 . Th e following are additional examples o f resolvingexpression s in to factors.

we (463e") (a:

3 —~

e3)

8b3 an s(2b)3 (so)3 (26 3 c) 26 x 3 0+

(2b 3 0) (4 b2 6bc

4 (ab+ cd)2—(a

2+ b2—c

2— d2)2=

— 02—d2)} —(a

2+ bz—c

’ —d 7)b2— c

z—d2} {2ab+ 2cd— a'2

{(a b)”(c cit } d)

2

(a W}—c+d) (a —a).

95 . Suppose that (x2— 5wy 6g”) (x—4y ) is to be divid

ed by fez7xy + 12y

2. We might multiply a

“

by x—4y , and then divide the result by x2 7xy + 12y

2.

But the form o f the question suggests to us to try ifx —4y is n o t a factor o f x2— 7xy + l 2y2 ; and we shall findthat x2 7my 12y

2= (a: 3y) (3: 4y). Then

(x2 5 xy 6y?) (x 4y) 5 xy 6g

g

(w—3e) (x — 4e) w—3y

and by divis ion we find thatx2 5 xy By

2

5 2 EXAMPLES . X I.

96. Th e studen t with a little practice wil l be able toresolve cert ain trin omials in to two bin omial facto rs.

For we have gen erally

suppose then we wish to kn ow if it be possible to resolvea? 7.v + 12 in to two bin omial factors we must find, ifpossible, two numbers such that their sum is 7 and theirproduct is 12 ; and we see that 3 and 4 are such numbers.

Thus

Similarly, by the aid of the f ormula(x - b)= a

3

we can resolve x”12 in to the factors (a: 3 ) (a:

And, by the aid of the formula

- b) : .v9+ (a —b) x - ab,

we can resolve x2 .v 12 in to the factors (a 4 ) (.c

We shall n ow give for exercise some miscellan eousexamples 111 the preceding Chapters.

EXAMPLES. XI.

Add together the following expressions :a (a + b—c), b (b+ c - a), c (a + c —b).

a (a b (b c (c

a (a b (a + b c (a + b+ c—d),d (

3a (4b 3b (4 c 7a ), 3 c (4a 7b).

9a 9b 9c

(b —c) y,

(c - a) y .

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5 4 EXA l l!PLES . X I.

D ivide25 .

26 .

27.

28.

29.

3 0.

3 1.

3 2.

3 3 .

3 4 .

3 5 .

3 6.

a6y6 2.v

3y3 by (x y)3.

x5y6 2ai’y

3 by (a:

(a33 a

gb 5 ab2 3b3) (a 2b) by a’3ab 2b’ .

9x2y 23 .ryg l 5y

3) (.v 7y ) by x2 Sp y

a8

a‘b“ b8 by (a

zab b?) (a

2ab

a8 b8 a

2b2 (a4 b“) by (a

2ab b”) (a

’ab

4 af’b2 2 (3a

4 2b4) ab(5 a2 1 1b2) by (3a b) (a b).

(x2

(cc— 3 ) by x2—5x + 6.

(x3—3x + 2) (x + 4 ) by x2+ x—2.

(a2+ aw+a 2) by a

4+

(a4

ai’b2 b“) (a b) by a

2ab b’.

—a2) by (a +b)

Resolve the following expression s mto factors3 7.

3 9 .

4 1.

4 3 .

4 5 .

4 7.

4 9.

5 0.

x2+ 9x + 20. 3 8. w2+ l l x + 30.

x2— 15 x + 5 0. 40. x

9

x9+zc — 132. 42. x

2—7x — 4 4 it t a, c

x4— 8 1. 44 . mu ms.

res—25 6. 4 6. a

s — 64 .

a’+ 9ab+ 20b2 4 8. x

2

(a b)’

1 1c (a b)+

Shew that the following resul ts ar e true :5 1. (a + 2b) a

s —b)

5 2. a (a —2b)3—b (b —b)

GREA TEST COMM ON M EAS URE . 5 5

XII. Gr ea test Common M easur e.

97. In Arithmetic a whole number which dividesan other whole number exactly is said to be a measur e o fit, o r to m easur e it ; a whole n umber which divides twoo r more whole numbers exactly is said to be a commonmeasur e of them.

In Algebra an expression which divides an other expression exactly is said to be a m easur e of it, or to m eas ur e

it ; an expression which divides two or more expressionsexactly is said to be a common measur e of them.

98 . In Ar ithmetic the gr ea test common measur e o ftwo o r more whole numbers is the greatest whole n umberwhich will measure them all . The term gr eatest commonmeasure is als o used in A lgebra, but here it is n o t veryappropriate

,because the term s gr ea ter an d less ar e sel

dom applicable to those algebraical expression s in whichdefin ite numerical values have n o t been assign ed to thevarious letters which occur. It would be better to speako f the h igh est common m easur e

,or of the h igh est common

divisor ; but in con formity with established usage weshall retain the term gr ea test common m easur e.

The letters G . C. M . will often be used fo r shortnessinstead of this term.

We have n owto explain in what senée the term is usedin Algebra.

99. It is usual to say, that by th e greatest commonmeasure o f two o r more simple expression s is mean t thegr ea test exp r ession wh ich wil l m easur e th em a ll ; but

this definition wil l n o t be fully understood un til we havegiven and exemplified the rule for finding the greatestcommon measure of simple expression s.Th e following is the Rule for finding the of

simple expressions . Fin d by A r ithmetic th e ofth e n umer ica l coefi cien ts ; af ter th is n umber p u t ever yletter wh ich is comm on to all th e exp r ession s , and giveto each letter r esp ectively th e least in dex wh ich it h as

in the exp r essions.

an other method, and we shall now give the uand rule .

103 . The followi ngmay be given as th e

the greatest common measure o f compoundLet two o r m or e comp oun d exp r ess io ns co n t

Qf some common le tter ; then th e f actor ofmen sio ns in th a t letter wh ich d ivides all theis cal led their gr ea test commo nmeasur e.

104 . Th e following is the Rule fo r findingcommon measure o f two compound

Let A a nd B den ote th e two

be a r r anged accor ding to des

common letter,a nd supp o se t

gr ea test

105 . Fo r 0

and 4x3—9x3— l 5 x -t

fin ition

”CG

ssion s

eatest

5 6 GREA TES T COMM ON MEASURE .

100. Fo r example ; required the G.C.M . o f 1 6a‘b2e and

20a3b3d. Here the numerical coefficien ts ar e 16 and 20

,

and their o n e. M. is 4 . The letters common to both theexpression s ar e a and b ; the least index o f a is 3

, and

the least index o f b is 2. Thus we obtain 4a3bzas the r e

quired G .C.M.

Again ; required the G.C.M . of 8a9b3c2x ’yz

3,12a‘bcx2y

3,

and 16a3e3x

2y4. Here the numerical coefficien ts ar e 8,

12, and 1 6 ; an d their G .C.M. is 4 . The letters common toall the expression s ar e a , e , x , and y ; and their least indicesar e respectively 2 , l , 2, and 1 . Thus we obtain 4 a2ex2y as

th e required G .O.M .

101 . Th e following statemen t gives the best practicaln otion of what is mean t by the term greatest commonmeasure, in Algebra, as it shews the sen se of the wordgr ea test here. Whe n two o r m or e exp r ession s a r e dividedby th eir gr eatest common measur e

,th e quotien ts h ave n o

common measur e.

Take the first example o f Ar t. 100, and divide the expression s by their the quotien ts ar e 4 ae and 5 bd

,

and these quotien ts have n o common measure.

Again , take the second example o f Ar t. 100, anddivide the expression s by their the quotien ts are

2b3cx3z3 , 3 a9by

2

,and 4 ac2y

3, and these quotien ts have n o

common measure.

102. Th e n otion which is supplied by the precedingArticle, with the aid of the Chapter on Factors, wil l enablethe studen t to determin e in many cases the G.C.M . o f com

p ou n d exp r ession s. Fo r example ; required“ the G .C.M . ofan d 6ab (a

2—b2). Here 2a is the G .C.M . o f th efactors 4 a2 and. 6ab; and a + b is a factor o f andof a

2—b2, and is the on ly common factor. Th e product2a (a b) is then the G.O .M. o f the given expression s.But this method cann ot be applied to complex ex

amples, because the gen eral theory o f the resolution o fexpression s in to factors is beyond the presen t stage o fthe studen t’s kn owledge ; it is therefore n ecessary to adopt

GREA TES T COMM ON M EAS URE . 5 7

an other method, and we shall now give the usual definitionand rule.

103 . Th e followingmay be given as the defin ition o f

the greatest common measure o f compound expressions .

Let two or m or e comp oun d exp r essio n s con ta i n p ower sof som e common letter ; th en th e f actor of h igh est dim en sions in th a t letter wh ich divides al l th e exp r ession sis cal led th eir gr eatest common measur e.

104 . The following is the Rul e fo r finding the greatestcommon measure o f two compound expression s.

Let A and B den ote th e two exp r ession s ; let th embe ar r anged accor ding to descending p ower s of som e

common letter,an d supp ose th e in dex of th e h ighest

p ower of th a t letter in A no t less th an th e in dex of theh igh est p ower of tha t le tter in B . D ivide A by B ;th en make th e r ema in der a diviso r an d B th e dividen d.

Aga in make th e n ew r ema in der a d ivisor a n d th e p r e

ceding divisor th e dividen d. Pr oceed in th is way un til

th er e is n o r emain der ; th en th e last divisor is th e

gr ea test common m easur e r equ ir ed.

105 . Fo r example ; required the G.C.M . o f x3 — 4x + 3

and 4x3—9x‘3 — l 5 x - l 18 .

—4x + 3 ) 4x3 9x2—1 5x + 18 ( 4x + 74x3 — 1 6x3 - i- 12x

7x 2— 27x + 18

7x 2—28x + 21

x — 3

x—3 ) x2—4x + 3 kx — I

x2—3x

Thus x —3 is the G.O.M . required.

GREA TES T COMM ON M EAS URE

106. Th e rule which is given in Ar t. 104 depends onthe following two prin ciples.

(1) If P measure A , it will measure mA. Fo r leta den ote the quotien t when A is divided by P ; thenA = aP ; therefore mA z maP ; therefore P measuresmA .

(2) If P measure A and B , it will measur e mA i n B .

Fo r , sin ce P measures A and B , we may suppose A aP,

and B = bP ; therefore mA thereforePmeasur es mA i nB .

107. We can n ow demon strate the rule which is givenin Ar t. 104.

Let A and B den ote the two ex

pression s. D ivide A by B ; let 1)

den ote the quotien t, and C the r e

mainder . D ivide B by C ; let q den ote the quotien t and D the remainder . D ivide C by D ,

and supposethat there is n o remainder

, and let rden ote the quotien t.

Thus we have the following results :A =p B + C, B n + D , O= r D .

We shall first shew that D is a common measure o fA and B . Because C= r D

, therefore D measures C ;therefore, by Art . 106, D measures 90, and als o qC+ D ;that is

,D measures B . Again , sin ce D measures B and C

,

it measures p B + C ; that is, D measur es A . Thus Dmeasur es A and B .

We have thus shewn that D is a common measure ofA and B ; we shall n ow sh ow that it is the ir gr eatestcommon measure.

By Ar t. 106 every common measure of A and B mea

sures A —p B ,

that is C ; thus every common measure ofA and B is a common measure of B and C. S imilarly,every common measure o f B a nd C is a common measure

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60 GREATES T COMM ON MEASURE .

x — l ) 2x’ — 7x + 5 (233—5

2x’—2x

Thus we obtain x—l as the required.

Here it will be seen ‘ that we used the second part ofthe rule of Ar t . 1 10, at the beginn ing o f the process , and

the first part o f the rule later. The first part of the rul eshould be used if possible ; and if n o t , the second part. Wehave used the word exp r ession in stating the rul e, but inthe examples which the studen t will have to solve

,the

factors in troduced or removed will be almost always n u

mer ica lf actor s, as they are in the preceding example.

We will n ow give an other example ; required the G .C.M .

of 2x4—7x 3 — 4x x —4 and 3x 4 — 2x2 — 4x 16 .

Multiply the latter expression by 2 and then take it fordividend.

2x4—7x3 — 4x’4- x—4 ) 6x‘ — 22x3 4x3 — 8x —32 (3

—21x3 — 1 2.v2+ 3x —12

{79+ 8x

’ — l l x — 20

We may multiply every term of this remainder by'

1

before using it as a n ew divisor ; that is, we may changethe sign o f every term.

x3 2x ‘

7x3 4x’+ x—4 tam e

40x

9x3 — 26x2 3 9x 4

9x 3—72x2 4 99x + 180

4 6x2 184

Here 4 6 is a factor of every term o f the remainder ; weremove it before using the remamder as a. n ew divisor.

EXAMPLES . X I] . 61

—3x—4) x3 —5

x3—3x2 4x

—5 x2+ 1 5x + 20- 5 x

2+ 1 5 x + 20

Thus .vz 3x 4 is the required.

1 12. Suppose the original expression s to con tain a

common factor E ,which is obvious o n in spection let

A : dF and bF. Then , by Ar t. 109, FWill be a fact01

o f the G .C.M . Find the G.O.M . of a and b, and multiply itby F; the product will be the G.C.M . o f A and B .

1 13 . We n ow p1 o ce ed to the G.CHM of more than twocompound exp1 e ssio n s. Suppose we 1 equir e the G. o f

th r ee expression s A ,B

,C. Fin d the of an y two o f

them,say of A an d B ; let D den ote this then the

G .C.M . of D and C will be the requ ired G .O .M . ofA,B

,andC.

For,by Art. 108, every common measure of D and C i s

a common measm e of A , B ,an d C ; and by A 1 t. 109 every

common measure of A ,B

,and C 1s a common measure o f

D and C. Th e ref01 e the o f D and C Is the G.O.M.

o f A,B , and C.

1 14 . In a similar mann er we may fin d the G.C.M . off our expression s. O r we may fin d the G.O.M . o f two ofthe given expression s, an d also the of the other two ;then the o f the two resul ts thus obtained will bethe G.C.M . of the four given expression s.

EXAMPLES. XII.

Find the greatest common measure in the followingexamples :

1 . 2. 1 6a2b3, 2oa3b2.

3 . 3 6x4y5z6,4 8x6y

5z4. 4 . 3 5 a

‘3b3x3y4,4 9a2b4x 4y

3.

5 . ate 6. see

LEAS T COMM ON M ULTIPLE .

XIII. Least Common .Multip le.

1 1 5 . In Arithmetic a whole n umber which is measuredby an other whole number is said to be a mul tiple o f it ; awhole number which is measured by two o r more wholenumbers is said to be a comm on mul tip le

'

o f them.

1 1 6 . In Arithmetic the least comm on mu ltip le o f twoo r more whole numbers is the least whole number whichis measured by them all . The term least common multipleis also used in A lgebra, but here it is n o t very appropriate ;see Art . 98 . The letters L.C.M . will often be used forshortn ess instead of this term .

We have n ow to explain in what sens e the term is usedin Algebra.

L

1 17 . It is usual to say, that by the least common multiple of two o r more simple expression s

,is mean t the least

exp r ession wh ich is m easur ed by th em a ll ; but this dehn itien will n o t be fully un derstood un til we have given andexempl ified the rule for finding the least common multipleo f Simple express ion s .

The following is the Rule fo r finding the L.C.M . o f

simple exp r ession s. Fin d by A r ithmetic th e L.C.M . of th e

n umer ica l coefficien ts ; af ter th is n umber p u t ever y letterwh ich occur s in th e exp r ession s, a n d give to each letter

r esp ectively th e gr ea test in dex wh ich it h as in the ex

p r essio n s.

1 18. For example ; required the L.C.M . o f 1 6a 4bc and20a3b3d . Here the numerical coefficien ts are 16 and 20,and their L.C.M . is 80. The letters which occur in the ex

pression s ar e a, b, c, and d ; an d their greatest indiceS '

ar e

respectively 4 , 3 , 1 , an d 1 . Thus we obtain 80a 4b3cd as therequired L.C.M .

Again ; required the L.C.M . o f 8a9b3c2x5yz

3,12a4bcx2y

3,

and 1 6a 3c3x9y4. Here the L.C.M . of the numerical coefficien ts

is 4 8 . The letters which occur in the expression s ar e

a,b,c,x, y , and z ; an d

tively 4 , 3 , 3 , 5 , 4 , and 3 .

the required L.C.M .

64 LEAS T COMM ON M ULTIPLE .

1 19 . Th e following statemen t gives the best practicaln otion of what is mean t by the term least common mul tiplein Algebra, as it shews the sen se of the word leas t here.Wh en th e lea st common m ultip le of two o r m or e exp r ession s is divided by th ose exp r ession s the quotien ts h ave n o

common measur e.

Take the first example o f Ar t. 1 18, and divide the L.C.M .

by the expression s ; the quotients ar e 5 b2d and 4 ae,and

these quotients have n o common measure.

Again ; take the second example o f Art . 1 18, and dividethe L.c.M . by the expression s ; the quotien ts ar e 6a

’cy

3,

4b2c2x3yz3,and 3ab3x3z3, and these quotien ts have n o com

mon measure.

120. Th e n otion which is supplied by the precedingArticle

,with the aid o f the Chapter o n Factors, will en able

the studen t to determ in e in man y cases the L.C.M . o f com

p oun d exp r ession s. For example,required the L.C.M . o f

4 a2(d + b)

2and 6ab (a 2—b

2

). The L.C.M . of 4a 2 and Gab is

12a‘3b. Also (a + b)2 and — b2 have the common factor

a + b,so that —b) is a mul tiple o f (a - t- b)

‘l

and o f a 2 b2 ; and o n dividing this by (a + b)2 and a2—b2 we

obtain the quotien ts a —b and a + b, which have n o commonmeasure. Thus we obtain 12a2b —b) as the required L.C.M.

121 . Th e following may be given as the defin itiou of:theL.C.M . of two or mor e comp ound exp r essions . Let twoor more compound expression s con tain powers o f somecommon letter ; then the expression of lowest dimens ion sin that letter which is measured by each of these expressions is called their least common multiple.

122. We shall n ow sh ow how to find the n ear. of twocompound expression s. Th e demon stration however willn o t be fully understood at the presen t stage of the studen t’skn owledge.Let A and B denote the two expressions, and D their

greatest common measure. Suppose A = aD,and B = bD .

Then from the natur e o f the greatest common measure, a

LEAST COMM ON M ULTIPLE . 65

and b have n o common factor, _and therefore their least

common multiple is ab. Hen ce the expression of lowestdimen sion s which is measured by aD and bD is abD . And

Hen ce we have the foll owing Rule for finding the L.e .M .

of two compound expression s. D ivide th e p r oduct of th e

exp r ession s by th eir G.O.M . O r we may give the rule thusD ivide o n e of th e exp r ession s by th eir G.O.M and mul

tip ly th e quotien t by th e oth er exp r ession .

123 . Fo r example ; required the L.C.M. o f x2 4x 3

and 4x3 9x2 1 5 x 18.

Th e G .e .M . is x 3 ; see Art . 105 . D ivide x2—4x 3

by x 3 ; the quotien t is x 1 . Therefore the L.C.M . is

(x 1 ) (4x3 9x3 1 5 x and this gives, by multiplying

o ut,4x4 1 3x 3 6x2 3 3x 18 .

It is however often conven ien t to have the L.e.M .

expressed in factors, rather than mul tiplied out . Wekn ow that the which is x — 3 , will measure the ex

pression 4x3 — 9x2—1 5 x - l- 18 ; by division we obtain thequotien t. Hen ce the L.C.M. is

(x — 3 ) (x —l ) (4x2+ 3x 6)

For an other example, suppose we require Le n .

2x2 7x + 5 and 3x 2—7x + 4 .

Th e G.c.M . is x 1 : see Art . 1 1 1 .

Also (2x2—7x + 5 ) —(x — 1)= 2x — 5

,

and (3x2 —1 ) : 3x —4 .

Hence the L.C.M. is

(x 1) (2x 5 ) (3x

.Again ; required the L.e.M . o f 2x4—lx3—4x2+x 4,

and 3x4—l 1x3—2x ’ —4x — 16 .

Th e G .C.M . is x2—3x—4 : see Art . 1 1 1 .

Also(2x4 — 7x3—4x

3+ x—4 ) —(x

2 3x - 4 ) 2x’x + 1,

and

(3x‘—1 1x3—2x2—4x—16) —(x9—3x —2x + 4 .

T. A. 5

66 LEAS T COAIM ON 111 ULTIPLE’

.

Hence the is

(x2—3x—4 ) (2x

2—x + 1) (3x2

124 . It is obvious that, every mul tip le of a commonmul tip le of two o r m or e exp r ession s is a common mul tip le

of th ose exp r ession s.1 25 . E very comm on mul tip le of two exp r ession s is a

multip le of th eir leas t common multip le.

Let A and B den ote the two expressions , bl theirand let N den ote any other common multiple. Sup

pose,if possible, that when N is divided by M there is a

remainder R ; let q den ote the quotien t. Thus R =N qMNowA and B measure M and N ,

an d therefore they measure R (Ar t. But by the nature o f divis ion R is oflower dimen sions than 111 and thus there is a commonmultiple o f A and B which is o f lower dimens ions thantheir L.C.M . This is absurd. Therefore there can be n o

remainder R ; that is,'

N is a mul tiple ofM1 26. Suppose n ow that we require the L.0.1 1. o f th r ee

compound expressions , A ,B , 0 . Find the o f any

two o f them,say o f A and B ; let M den ote this

then the of M and C will be the required o f

A ,B

,and C.

Fo r every common multiple o f M and C is a commonmul tiple o f A , B , and C, by A r t . 124 . And every commonmultiple o f A an d B is a mul tiple o f .M

,by A rt. 125 ; hen ce

every common multiple o f M and C is a common multipleo f A

,B , and C. Therefore the L.0.11. of M and C is theo f A ,

B,and C.

127. In a similar manner we may find the o f fourexpressions.

128. Th e theo ries o f the greatest common measure ando f the least common multiple ar e n o t n ecessary for thesubsequen t Chapters o f the presen t work, and any didi

culties which the studen t may find in them may be postpon ed un til he h as read the Theory of Equations . Th eexamples however attached to the preceding Chapter andto the resen t Chapter should be care fully worked , o n s e

coun t 0 the exercise which they afford in all th e funda»men tal processes of Algebra .

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68 FRACTIONS.

XIV. Fr action s.

129. In this Chapter and the following four Chapterswe shall treat o f Fractions ; and the studen t will find thatthe rules an d demon strations closely resemble those withwhich he is al ready familiar in Arithmetic.

a1 30. By the expression 5 we indicate that a unit is

to be divided in to b equal parts, and that a of such partsare to be taken . Here 23 is called a f r action ; a is calledthe n umer a to r

,and b is called the den om in a tor . Thus

the den om inator indicates in to how many equal parts theun it is to be divided, and the numerator indicates howmany o f those parts ar e to be taken .

Every in teger o r in tegral expression may be cons ideredas a fraction Wi th un ity for i ts den omin ator ; that is, for

b+ cexample,

b c1

1 3 1 . In Algebra, as in Arithmetic,it is usual to give

the following Rule for expressing a fraction as a mixedquan tity : D ivide th e n umer a tor by th e den omin a tor

,as

f ar as p ossible, an d an n ex to th e quotien t a f r ac tionh aving th e r emain der f o r n umer ator , an d th e d ivisor f o rden omin ator .

Examples.

- a +

x'Z— 3x + 4 x

i — 3x + 4

=x + 3

FRACTIONS. 69

Th e studen t is recommended to pay p ar ticula r a tten

tion to the last step ; it is really an example of the use ofbrackets

,namely, x (x

1 32. Rule for multiplying a fraction by an in teger.E ith er multip ly th e n umer ator by that int eger , o r divideth e den om in a tor by th a t in teger .

aLet

5

willgx c=a

g

e. For in each o f the fraction sgand the

un it is divided in to b equal par ts and c times as many0 as O a 0 0parts are taken 1n as m hen ce 13 c times ab b b b

This demon strates the first form o f the Rule.

den ote any fraction , and c any integer ; then

Again ; let den ote any fraction , and c any in teger ;( 6then willb?

x c For in each o f the fractions

and a the same number o f parts is taken , but each part

13 0 times as large as each part in because in

the un it is divided in to 0 times as many par ts as in

a 0 O ahence 18 0 t imes13 ; b

This demon strates the second form o f the Rule.133 . Rule fo r dividing a fraction by an in teger.

.

E ith er

multip ly th e den om in a tor by th at in teger , o r divide th e

n umer ator by th a t in teger .

Let 5; den ote any fraction , and c any in teger ; then

5 is 0 times by Art . 132 ; and

therefore i '

s1th o f

a.

be c b

This demon strates th e first form o f the Rule.

70 FRACTIONS .

Again ; let denote any fraction , and c any in teger

then will is 0 times 95, by Art . 132 ;and therefore 18 —th of

This demonstrates the second form o f the Rule.

1 34 . If th e n umer ator an d den omin ator of an y f r action be mu ltip lied by th e same in teger , th e value of th e

f r action is n ot a lter ed.

Fo r if the numerator o f a fraction be mul tiplied by anyin teger, the fraction will be multip lied by that in teger ;and the resul t will be divided by that in teger if its den ominator be multiplied by that in teger. But if we multiplyany number by an in teger, and then divide the result bythe same in teger, the number is n o t altered.

Th e result may also be stated thus : if the numeratorand den ominator o f any fr action be divided by the samein teger, the value o f the fraction is n o t altered.

Both these verbal statemen ts are in cluded in the algobraical statemen t a a c

b be

This result is o f very great importan ce ; many o f theoperation s in Fraction s depend on it

,as we shall see in the

n ext two Chapters.

1 3 5 . Th e demon strations given in this Chaptersatisfactory on ly when every letter den otes some p osi tivewh ole n umber ; but the resul ts ar e assumed to be truewhatever the letters denote. For the gr01mds o f thisassumption the studen t may hereafter con sul t the largerAlgebra. Th e result con tain ed in Art . 1 3 4 is the mostimportan t ; the student will therefore observe that hen ceforth we assume that it is always true in Algebra thata ac

b_

bc

aFor example, if we put 1 for c we have

5

whatever a, b, and 0 may denote.

EXAMPLES. XIV 71

9b

we obtain such results as the following

In like mann er,by assum ing that x c is a lways equal

x —l — 2

EXAMPLES. XIV.

Express the foll owing fraction s as m ixed quan tities25 x

7

x2+ 3x + 2

“x2— 6x — l

x 3

x3

ax2 3 a

”x 3a 3

Multiply2

3%by 3b. by 3 (Cl —b)

3 (a b)s (a

3+ b3)

by 4 (a ab+b ).(x

2- 1)

2by x + l .

D ivide8x 2

3yby 3a 2b.

10 (as—b3) 2

3by 5 (a +ab

by x"—x + 1.

72 RED UCTION o r FRAo r rozvs.

XV. Reduction of Fr actions.

13 6 . Th e result contain ed in Art . 1 3 4 will n ow be

appli ed to two importan t operations , the reduction o f a

fraction to its lowest terms, and the reduction o f fractionsto a common den om in ator.

13 7. Rule fo r reducing a fraction to its lowest terms.D ivide th e n umer a tor an d den om in ator of th e f r actionby th eir gr eatest common measur e.

For example ; reduce to its lowest terms.

Th e G .0 . 11 . of the numerator and the den omin ator is4a3b’ dividing both n umerator and denomin ator by 4 a 3b’,we obtain for the requir ed resul t

4 ae That is, 662 isequal to

1 6a4b?c

but it is ex r essed in a more simple20a 3b3d

’ P

form ; and it is said to be in the lowest terms , because itcann ot be further simplified by the aid o f Art . 13 4 .

x2 4x 3

Again l educe4x 3 9x2 1 5 x 1 8

to its lowest terms.

The of the mun erato r and th e den ominator isx 3 dividing both numerator an d denominator by x 3

x lwe obtain for the requl red result4x

33x 6

In some examples we may perceive that the numeratorand den omin ator have a common factor, m thout usmg therule for finding the Thus

,for example

,

(a —b)"

- c2(a —b—c) a —b+ c

az—(b -f- c)

z — b

RED UCTION OF FRACTIONS. 73

138. Rule for reducing fractions to a common denominator. M u l tip ly th e n umer a tor of each f r action bya ll th e den om in ator s excep t its own , f o r th e n umer a torco r r esp on ding to th a t f r action ; an d m ultip ly a ll th eden omin a to r s togeth er f o r th e common den om in a tor .

eFo r example ; reduce and to a common de

f

c cbf e ebd

b_

bdj"

d_

dbf’

f f bd'

nominator.

gi

g“

; dif

f" are fractions o f the same

val ue respectively as and they have theb Et ’

common den ominator bdf .

Th e Rule given in this Article will always reduce fraction s to a common den ominator

,but n o t always to the

lowest common den om inator ; it is therefore often con

venien t to employ another Rule which we shall n owgive.

- 139. Rule for reducing fractions to their lowest common den omin ator. Fin d th e least common mul tip le ofth e den om in a to r s, an d take th is f o r th e commo n den omi

n a tor ; th en f o r th e n ew n umer a to r cor r esp on ding to an yof th e p r op osed f r a ction s, multip ly th e n umer a to r of th a t

f r action by th e quo tien t wh ich is obtain ed by dividingthe least common mul tip le by th e den omin ator of th at

f r action .

Fo r example ; reduce £5 b i to the lowest comya xx xy

mon den ominator. Th e least common multiple o f the den ominators is xyz ; and

74 EXAMPLES. XV

EXAMPLES.

10a2x

5 a2x 1 5 ay2

2x2+ x — 1 5

2x2— 1 9x + 3 5

xz

x2+ (c—a) x — ac

(x + b)2

x2— 10x + 21

x3—4 6x — 21

x2+ x — 42

x3

20x2+ x — 12

12x 3—5 x2 + 5 x — 6°

2x3 —5 x2— 8x—1 6

2x3 + 1 1x 3 + 1 6x + 16

x3—8x—3

x4

x 3 — x 9—7x 4- 3

x4+ 2x 3 + 2x — 1

XV

Reduce the following fractions to their lowest terms

x2+ 10x + 21

x2— 2x — l 5

-l 3 6

4x23 3x 27

x2+ 9x + 20

6x3 —l 1x + 5

3x 3—2x“ 1

x2 2ax a

2

x 3 2ax2 2a2x a3

x3 3 e

’x 2a 3

2x 3 ax’

d’x —4 a3

x3

as

x4

anzi a“

3x4— l ix 3 —9x + 2

2x4 — 9x3—14x -r 3'

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76 AD D ] TION OR SUB TRACT] ON

XVI. Addition o r Subtr action of Fr actions .

140. Rule for the addition o r subtraction of fraction s. Reduce th e f r action s to a comm on den om ina torth en add o r subtr act th e n umer a tor s and r etain the comm on den omin a tor .

Examples. Add

Here the f1 act10n s have al ready a common den ominator,

and ther ef01 e do n o t require reducinga + o a —c a + c + a—c 2a

b b b b

3 a — 4b

c

4 a—3b 3 a —4b 4a — 3b— (3a—4 b)0 c c

4 a — 3b—3 a + 4b a + b

c c

Th e studen t is recommended to put down the work a t

f ull , as we have don e in this example, in order to ensureaccuracy.

c c

a + b a —b'

Here the common den ominator will be the product o fa +b and a—b, that is aZ— bi

c c (d —b).

c c (a -l—b)a + b a

2—'

b2 a —b a‘z—b

’3

a + hTh erefore c (a 2

4

65ca — cb+ cd + cb 2ca

a2

_ b2—a2

_ b2 °

OF FRACTIONS .

From

Th e common den ominator is a2—b2.

a + b (a + b) a—b (a—b)a—b a

2—b2 ’a + b a

2— b2°

a +b a—b (a + b)z— (a—b)

2

Thereforea b a b a

2 b2

a2+ 2ab+ b2—(a 2 - 2ab - l- b?) 4ab

ail —b2 a

2—b2'

4x2 - 3x + 2From4x

3—9x‘3— 1 5 x + 18

'

By Art . 123 the o f the den omin ators is

(x 1 ) (x 3) (4x2+ 3x

x + l —6)x2—4 x + 3 (x

4x2—3x 4—2 (4x2 - 1)

4x 3 — 9x2—1 5 x -1 18 (x—1) (x

x + 1 4x2— 3x + 2Therefore

x2—4x 3 4 x ~i 9x 2 1 5 x 18

(x + 1) (4x2+ 3x — 6 )— (4x

2 —1)(x —1 ) (x —6 )

4x 3 + 7x3 —3x—6 - (4x3 — 7x

2+ 5 x—2)

(x—1 ) (x —6)

14x2— 8x—4

(x 1) (x 6 )

77

78 AD D I TION OR S UB TRACTION

14 1. We have sometimes to r educe a m ixed quan tifyto a f r action ; this is a simple case of addition or subtraction o i fraction s.

b a b ac b ac+ b

c l c c c c

2ab a+2ab a (a + b) 2ab a

2+ 3 ab

1

Examples.

a + bz

a +b a + b a + b a + b

x—2_

x —2

x2—3x + 4

_

xi — 3x + 4

—3x + 4 )x2—3x + 4

x3 — 5 x + 12—(x ;—2) x

3—5 x + 12 - x + 2 x3 —6x + 1 4

x2— 3x + 4 x

2— 3x + 4 x2—3x + 4

1 42. Expressions may occur involving both additionand subtraction . Thus

,for example

,simplify

a ab de

a + b a"3

- b2 a2+ b2

°

Th e o f the den ominators is (a2

that is a 4 — b‘

a (a b) (a2 b?) a

4aab a

'-’

b2 ab3

a4 b‘ a

t ( fl

ab (a’ b’ ) a

3b ab3

a“’—b2 a

4—b* ai —b4

a2(a2 b?) a

‘a’zb’

a“ b“ a

4 b4

a a2

Thereforea + b a

2 b2 a? b3

a4

a3b a

zb2 ab3 a3b ab3 (a‘ a

‘b ”)

a4

_ b~l

a4—a3b+ a

?b2— ab3 + a3b+ ab3 —a

‘+ a

263

ai

_ b1 as bl

‘

OF FRAcrzozvs. 79

a b 0

8mm“? “

(a —b) ("

J —7 ) <b— c) (If—a )4“

<c—a) (c— br

Th e begin n er shoul d pay particul ar atten tion to thisexample. He is very liable to take the product o f theden om inators for the common den omin ator, and thus torender the operation s extremely laborious .

The second fraction con tain s the factor b a in its de

n om in ator,and this factor differs from the factor a b,

which occurs in the den om in ator o f the first fraction , on lyin the sign o f each term , and by Art . 1 3 5 ,

b

(b—Cl (b—a ) (b—c) (d —b)'

Also the den ominator o f the third fraction can be putin a form which is more conven ien t for our object ; fo r

pby

the Rule of S ign s we have(c—a ) (c —c) (b—c).

Hen ce the proposed expression may be put in the forma b c

(a—b) (a — c) (b— c) (a —c) (b— c)’

and in this form we see at on ce that the o f the denomin ators is (a b) (a c) (b o).

By reducing the fraction s to the lowest common den ominato r the proposed expression becomes

a (b— c) - b (a —b)

(a—b) (a—c) (b—c)

that isab— ac— ab+ bc + ac - bc

, that is 0.

(a—b) (a—c) (b— c)

14 3 . In this Chapter we have shewn how to combin etwo or more fraction s in to a single fraction ; on the otherhand we may, if we please, break up a single fraction in totwo o r more fractions. For example

,

3 be— 4ac+ 5 ab 3be 4 ao 5 ab 3 4 5

abc abc abc a b

80 EXAMPLES . X VI .

EXAMPLES. XVI.

Find the value o f

3 a—5 b 2a— b— c a + b+ c

l —3x 1 + 3x'

261—26 26—2a°

a

a —x a + x a —x 2

a — 2b b— 3c 4 ab+ 3bc

6ao

a —b+

2a a3 + a

gh

b a — b a‘ib—b‘“

2b b— 2a+3x (a —b)

x x + b .t’2— b3

3 5 2x—7x 2x—1 4x

2— l'

( 1+

b

EXAMPLES. X VI.

1 3 2x

x —2 x + 2

1 1 a

a —b a + b a2— b2

'

a + x+a —x a

”'—x2

a —x d + x

1 2 l

x + 1

x x

x— l x + 1 x - 2'

4x x —y+x + y

y x +v w—y'

x2

x2 1 . x

x—l x + l

x’

x2. 02 x

x + l+x—1

1 1 2

x —a x + x x

a a+4a2b2

a—b a + b a 4—b"

x’

x x

x2—l

+x—1

+x + 1

a 3a 2ax

a —x d + x a2+ x

2°

1 x + 10

2x — 4 x + 2 2x 2 + 8°

2 x—3

x + 4 xz

1 1 l

x2 —iz)

2'

T. A.

81

82 EXAMPLES X VI .

x2+ ax + a

2.x2

x3—a

3x3+ a

3

y2

xy + y2

x2+ xy

'

x2— 2x + 3 x — 2 1

o

x5+ 1 —x + l x + l

l 2x—3 1

x (x + 1)

1 — 2x x + 1 1

3 (x2—x + l )

l x

2- l

y

xz—xy + y x + y x

3+ y

3°

l x —y

x —y x

2+ xy + y

2

x + x + l xg—x + 1 x

4+ x

1

+ l

a + b a—b

o x + by ax —by a'3x2+ b2y

2

l 1

xi—x

'i—l—l .v

z

1 2 3

x2 —7x + l 2 a

'2—4x + 3 x2—5 x + 4

1 1

2

l 1 2b

a —b a + b aP

-r b2 a4+ b4

°

1 1 3 3

x—3a x + 3a x + a x—a

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84 M ULTIPLICA TION OF FRACTIONS .

XVII. .Mul tip lica tion of Fr action s .

1 4 4 . Rule for the mul tiplication o f fractions . M ulti

p ly togeth er th e n um er a tor s f o r a n ew n umer a tor , andth e den om in a tor sf o r a n ew den om in a tor .

14 5 . The following is the usual demons tration o f the

Rul e. Leta

and f be two fr actions which are to bed

mul tiplied together ; put ; —x , and y ; therefore

a s ,and c z dy ;

therefore ac bdxy

divide by bd, thus g§=xux

‘2“

b d

a c actherefore

bx

d bd'

And ac is the product o f the numerators, and bd theproduct o f the den omin ators ; this demonstrates the Rul e.Similarly the Rul e may be demon strated when more

than two fractions are mul tiplied together.1 46. We shall n ow give some examples . Before multi

plying together the factors o f the n ew numerator and t h efactors o f the n ew den om in ator, it is advisable to examin eif any factor occurs in both the numerator and den ominator, as it may be struck out o f both, and the result willthus be simplified see Art. 137.

Multiply a by

_ _

a.

axb ab

‘

1 1 c c

Hence asand are equivalent ; so, for example,

2x — 3loand

4( - x 3 )

4

M ULTIPLICA TION OF FRA o rzozvs . 85

Multi 1 i”bp ry

yy

xxx x x x x

2

y y w e ez’

Multiply

3ax80 3 a x 80 2c ><12a 2c

4b 9a 4b x 9a 3 b><12a 3b'

3 a”

Mul tiply(a + b)

2by

4 (a2—b2) 4a (a—b) x 3a (a + b)3 ab

a b a bMul t1ply 5

+a+ 1 by b

+a

a2+ b2+ ab

ab ab ab ab

a? b2 ab a

9+ b

2 —ab.

amt ab ab

a2+ b2+ ab a

2+ b2— a b (a2+ b

' — ab)ab

(a‘Z

- l- bZP— ai’b2 a

“+ b“ - a

?b2

afib‘

O r we may proceed thus :b a b

a

86 M ULTIPLICA TION OF FRACTIONS .

therefore

Th e two results agree, for 5 72 2+ 1

1 — a2 1 b

a

d

b b2 a a”

We might mul tiply together the first two factors,and

ab

1 — a’

add the results ; but it is more conven ien t to reduce themixed quantity b+

ab to a single fraction . Thus

ab b (1 — a) + ab 6

Multiply together and b

then multiply the product separately by b and by

l —al—bz

><b (1 —a

2) ( 1 —b

'

b+ b2 a + a2><l —a —a )

147. As we have al ready don e 111 former Chapters, wemust here give some results which the studen t must assum e to be capable of explan ation , and which he must us eas rules in working examples wh ich may be proposed. See

A rts. 6 3 and 1 3 5 .

Multiply gby

Multiplya C — ( l 0 acX ~ x

b d b d bd

Mul tiply

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88 D I VISION OF FRACTIONS .

XVIII. D ivision of Fr action s.

148 . Rule for dividing on e fraction by an other . In ver tth e divisor an d p r oceed as in M ultip lication .

149 . Th e following is the usual demon stration o f thea a

Rule. Suppose we have to dividebby 2, put b x,

and20

3: y ; therefore

a = bx, and c= dy ;

therefore ad bdx, and be bdy

ad bdx x

thereforebe bdy y

x a

tl e ef r6} 0 ad a

xd

1 1‘

0 9b

'

d—EU b c

a

1 5 0. We shall now give some examples.

D ivide

3 a 9a 3a 8c 2c><1 2a 2c

4 b'

8c Exact 3b x l 2a sb

'

a b—b“’D m de

(a b)2by

a2 bz

'

( lb—b” ab— b2 aQ—b2

a2

x

b2

D I VISION OF FRACTIONS 89

1 5 1 . Complex fractional expression s may be simplifiedby the aid o f some o r all o f the rul es respecting fractionswhich have n ow been given . Th e following ar e examples .

a — b [ a + b a —bS‘mph fy

a + b'

ta—b a + h

a + b+a—b —b)2

a—b a + b (a aZ—b2

’

a +b a —b (a + b)z—(a—b)2 4ab

a —b a +b (a a2—b2

’

2ab - 2b2 4 ab 2a2 - l- 2b“’xa2— b2 a

"‘- 1—b2

—b2'

a2—b2 a

2 - b2 4ab 2ab

In this example the factors a b and a +b aremu lti

p lied togeth er , and the result a” b2 is used in stead o f

(a + b) (a b) ; in gen eral however the studen t will find i tadvisable n o t to mu ltip ly th e factor s together in thecourse o f the operation , because an opportun ity may occuro f striking out a common factor from the numerator an dden ominator o f h is result.

a + 1 3 — a+a + 1 3 — a + a + 1

3 — a 3 —a 3 — a

4 1X3 —a 3 — a

3 — a 1 4 4

3 —a 4a+3—a 3 + 3a

4 4 4 4

3 + 3a 1 4

4 1x

3 + 3a 3 + 3a'

90 D I VISION OF FRACTIONS .

Find the value o f when x

2ab a 2ab ab—a2

2x —a =a + b 1 a + b

2 —b2ab b 2ab ab —b2

xa + b 1 a + b a + b

'

Therefore 2x—a ab— a

2ab—b2 ab—a2

xa + b

2x —b a + b a + b a + b ab —b2

ab— a2

a (b— a) a

ab— b2_

b (a - b) b’

therefore

a ab a a + b - ab a2

Again , a — x z

i a + b a + b

_

b ab b (a + b) - ab b”“

1 a + b a + b a + b'

a —x b” a + b a“

Thereforeb—x a + b a + b a + b

x

b2 b”'

2 2

Therefore a x

Z”

,

(

22b—x

1 5 2. Th e resul ts given in Art . 14 7 must be givenagain here in conn exion with D ivis ion of Fractions.

h0 awe ave

bd.

d—b bd

.

d b.

x0 ac

we haved—bd ’Also since

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EXAMPLES . 11VIII.

by x -

x

x2 1 x

‘

1 1

x by 52 g4 “

lb

a2

x

15 2by

13 1 +

a :v a x

Slmph fy the foll owmo'expr esswn s

3x+x—1

x — 1 +2 3

x—2 +

2x—1 x — a

tar—b) (as—c)

"

x a

25 . 1 26 l —F

1 x

x

2x2

1 + x +l —x

EXAMPLES . XVIII.

1

—b2 a2+ b

Find th e values of the following expressionsa—x when x

x — a x—bwb n

b ae x —

a —b'

{8 x a

a+b— a a + b

when x

when a

2

x +y x —y x2—ywh en y

x + 2a x—2a 4 ab

2b—x+2b+ x 4 b2—x2

When

3 x —2a + b when x =x + a—2b

a + 1and _

ab+ a

ab+ 1’ y _

ab+ 1°when x

93

941 SIMPLE EQUA TIONS

XIX. S imp le Equa tions

1 5 3 . When two algebraical expressions are conn ectedby the sign o f equality the whole is called an equation .

The expression s thus con n ected ar e called s ides o f theequation or m ember s o f the equation . The express ion tothe left o f the sign o f equality is call ed the fir st side

,and

the expression to the right is called the secon d side.

1 5 4 . An iden tica l equa tion is o n e in which the twosides are equal whatever n umbers the letters represen t ;for example

,the foll owing ar e

- iden tical equation s,

(x + a) (x —a)= x2—a

2,

(x a)zz x

" 2xx+a3

,

(x a) (x2—xa + a?

)= x3+ a 3 ;

that is,these algebraical statemen ts ar e true whatever

numbers x and a may represen t. The studen t will seethat up to the presen t poin t he h as been almost exclus ivelyoccupied with results o f this kind, that is, with iden ticalequations.

An iden tical equation is called briefly an iden tity .

1 5 5 . An equa tion of condition is on e which is n o t truewhatever n umbers the letters represent

,but on ly when

the letters represen t some particular number o r numbers.For example

,x + 1 = 7 cann ot be true un less x = 6 . An

equation of condition is called briefly an equa tion .

1 5 6 . A letter to which a particular value o r valuesmust be given in order that the statemen t con tain ed in an

equation may be true, is call ed an u nkn own qua n tity . Suchparticular value o f the unkn own quan tity is said to sa tisf yth e equa tion , and is called a r oot of the equa tion . Tosolve an equation is to find the root or roots.

1 5 7 . An equation involving on e unknown quan tity issaid to be o f as man y dimen sion s as the index o f thehighest power o f the unkn own quan tity. Thus

,if x den ote

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96 SIMPLE EQUA TIONS .

In stead o f mul tiplying every term by 3 x 4 x 6, we maymul tiply every term by 12, which is the L.C.M. o f the den ominato rs 3 , 4 , and 6 we shoul d then obtain at once

4x + 3x + 2x = 108

that is,divide both sides by 9 therefore

Thus 12 is the r oot o f the proposed equation. We may“

ver ify this by putting 12 fo r x in th e original equation.

The first side becomes1 2 12

3 4 6

which agrees with the second side.

that is that is 9 ;

1 6 1 . An y term may be tr an sp osed f r om one

an equa tion to th e oth er side by ch anging its sign .

Suppose, fo r example, that x a b y .

Add a to each side thenx—a + a = b—y + a,

that is x z b—y +a .

Subtract b from each side thusx—b= b+ a—y —b : a—y .

Here we see that —a has been removed from o ne sideo f the equation , and appears as a o n the other side ; andb has been removed from o ne side and appears as —b onthe other side.

162. If th e sign of every term Qf an equation bechanged the equa lity still h olds.

This follows from Ar t. 1 6 1 , by transposing every term.

Thus suppose, fo r example , that x—a = b—y .

SIMPLE EQUATIONS 97

By transposition y b a x,

that is, a—x = y—b ;

and this resul t is what we shall obtain if we change thesign o f every term in the origin al equation .

1 63 . We can n ow give a Rule for the solution of anysimple equation with o n e unkn own quan tity. Clear th e

equa tion of f r action s, if n ecessa r y ; tr an sp ose a ll th e

terms wh ich in volve th e un kn own qua n tity to on e side ofthe equa tion ,

an d th e kn own quan tities to th e other side ;divide bo th s ides by th e coefi cien t

,o r th e sum of th e co

efi cien ts, of th e un kn own quan tity , and the r oot r equir edis obtain ed.

1 64 . We shall n ow give some examples.

Solve 7x 25 3 5 5 x .

Here there are n o fractions by transpo sing we have7x — 5 x = 3 5 —25 ;

2x 10

divide by 2 ; therefore

We may verify this result by putting 5 fo r x in theoriginal equation ; then each side is equal to 60.

1 6 5 . Solve 4 (3x—2)— 2 (4x—3 )Perform the mul tiplication s in dicated ; thus

l 2x — 8 — (8x—6)

Remove the brackets ; thusl 2x—8—8x + 6 12 + 3x = 0 ;

collect the terms,tran spose

,

divide by 7,

Th e studen t will find it a useful exercise to ver ify theco rrectn ess o f his solutions. Thus in the above example,

ELL 7

98 SIMPLE EQUA TIONS .

if we put 2 for x in the origin al equation we shall obtain1 6 — 10— 6, that is 0, as it should be .

1 66. Solve x —2— (2x — 3)

Remove the brackets ; thus

x 2 2x 3

that is,

1 x

multiply by 2, 2 2x 3x l

tran spose,

2 1 2x 3x

that is, 1 5 x,o r 5 x 1

1fther e ore

5

1 67. Solve- 4 1 56 4- 5

2

the L.C.M . o f the den omin ators is 10 ;

5 (5 x + 4 ) x 2—5 (x

25 x + 20—7x — 5 x + 5 ;

25x

23x ='

4 6 ;

4 6

23

Th e beginn er is recommended to put down al l the workat full , as in this example, in order to ensure accuracy.

Mistakes with respect to the signs ar e often made in clearing an equat ion o f fraction s. In the above equation thefraction h as to be multiplied by 10, and it is ad

10

visable to put the resul t first in the form and

afterwards in the form —7x—5 , in order to secure attention to the signs.

therefore

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A 5 0.

EXAMPLES . X IX .

x + 3 x + 4+x + 5

2 3

7x + 9 2x —1

4— 7 + x

9

3x—4 6x — 5 3x—l

2 8

2x—5 5 x—3

3 4

x - 3 x —5 x—l

4"

6

x— l x—3+x — 5 x

2 4 6 1 m 3

7x + 5i

5 x + 6 8—5 x

6 4 12

x + 4 x—4 3x ; 1

+ 2§

x —l 2x + 7 x + 2

9

v—l x — 2+x—3 2 .

2 3 4 3

2x—5 6x + 3

6

(3x —5x.

7

“73+2 w

as x

_

1+ —75 . 5 2.w&

101

102 EXAMPLES . X IX .

5 —3x+5 x 3 3—5 x

4 3 2 3

5 4 . 327—27 3 (755

5 5 . 5 x—[8x 3 {16—6x 6.

1

'

- 2x 4 —5 x 1 3

3 6 42

x + l x — l_ 1. z

3 4M7 12+

7— 4x

5 x—1 9x — 5 9x—7

7 5

x + 3 x — 2 3x

2 3 4°

—x ) +x — 1 2

3x — l 13 —x 7x 1 1

5 2 s‘

e

2x — 1+6x —4 7x + l 2

5 7 1 1

2—x 3 —x 4 —x 5 x 3

3 4 5 6

—Q

5 43—3 9—x 194.

7 3 2 6<x 4)'

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1041 SIMPLE EQUA TIONS .

that is

+ 48x + 16x —165 + 64x o

By transposition and reduction

multiply by 15 —2x ; thus

l 44x —2x)= 60 8x ;

therefore I44x 8x 320 60,

that is, 1 5 2x = 3 80

therefore 1

1

3

22 25 .

x—5 x + 3173 . Solve

x 7 x 9

Multiply by (x 7) (x 9) thus

(x + 9) (x

that is, x2+ 4 .v—4 5 = x2— 4x

subtract x2 from each side of the equation , thus

4x 4 5 4x 21

transpose,

4x 4x 4 5 21,

that is, 24 ;

therefore

It will be seen that in this example is found o n bothsides o f the equation , after we have cleared o f fractions ;accordingly it can be removed by subtraction , and so theequation remains a simp le equation .

SIMPLE EQUA TIONS . 105

2x + 3 4x + 5 3x +'

3174 ‘ SOlve

x + 1 4x + 4 3x + 1

Here it is conven i en t to mul tiply by 4x + 4 , that is by4 (x

therefore 8x 12 4x 5

4x + 7

Mul tiply by 3x 1 ; thus (3x + 12 (x +

12x2+ 25x + 7= 12x2+ 24x + 12.

Subtract 12x2 from each side, and transpose ; thus25 x—24x = 12—7,

that is,

x— l x - 2 (x—1 ) (x—3 )— (xx—2 x — 3 (x—2) (x — 3 )

x2— 4x + 3 — (x

2—4x + 4 ) 1

(x—2) (x —3 ) (x — 2) (x

x — 4 x — 5 (x — 4 ) (x - 6)— (xx —5 x — 6 (x—5 ) (x - 6)

x2— l ox + 24 —(x2— l ox + 25 ) 1

(x — 5 ) (x - 6) (x

Thus the proposed equation becomes

106 SIMPLE EQUA TIONS .

l l

(x - 3 ) (x—5 ) (x - 6)

Clear o f fractions thus (x 5 ) (x 6) (x 2) (x

Change the signs ; thus

that is, x2—1 1x + 3 0 : x

2— 5x + 6 ;

therefore 1 l x 5 x : 6 30

that is 6x

6x : 24

therefore x z 4 .

°

4 5x°

3x1 76. Solve 5x + '

6.

2.

9

To en sure accuracy it is advisable to expres s all thedecimals as common fractions thus

10x12 10

2 10 9

Simplifying,

that is, 2+9

2

2:

2:

Mul tiply by 12, 6x 9x 15 72 4x 8

tran spo se, 19x = 72 + 8 4»

therefore

1 77. Equations may be proposed in which letter s are

used to represen t kn own quan tities ; we shall con tinue tore resen t the unkn own quan tity by x , and any other letterwi 1be supposed to represen t a kn own quan tity. We willso lve three such equations.

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108 EXAMPLES. XX .

181 . Although the foll owing equation does n ot strictlybelong to the presen t Chapter we give it as there will ben o difficulty in following the steps o f the solution , and itwill serve as a model for similar examples . The equationresembles those already solved

,in the circumstan ce that

we obtain on ly a single value o f the unkn own quan tity.

By tran sposition , J(x —,Jx

square both sides ; thus x JxY= 64 16Jx +

therefore 16 64 16 Jx ;

16

J¢ = 5 ;

therefore x = 25 .

EXAMPLES. XX.

29 3 5

x 12x 24°

x —2 x—3'

4 5 5 7

3x —4 5 x —6 2x + 3 4x—5'

3x—1 2x — 5+x — 3 x

2 3 4

x —x IO—x

4 6

5 x — 8 3x—8 x—2 1

1 10 EXAMPLES. XX .

(x—l ) (x—2) (x 3 (4x—2) (x +

(x—9) (x — 7) (x - 5 ) (x—1)—(x —2) (x - 4 ) (x—6) (x

(se - av(x —1)2(4x

xz—x + 1 x

2+x + 1

x + 1

3 8.

‘

5 2: 2‘

2x l .

2x.

3 9. 5 2:‘

6x‘

8‘

75x‘

25 .

'13 5x

'3 6

‘

09x40.

‘

1 5x +‘

6 2 9

4 1. aa '

b

'xbb”

42. aa

xz—a

" a—x 2x a

b b x'

4 4 . x (x —a) (x—b).

4 5 . (.v - a) (x - b) (x + 2a -l- 2b)

(x +2a) (x +2b) (.v—a—b).

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1 12 PROBLEMS.

XXI. Pr oblems.

182. We shall n ow apply the methods explain ed in thepreceding two Chapters to the solution of some problems

,

and thus exhibit to the studen t specimen s of the use o f

Algebra. In these problems certain quan tities ar e givenan d an other, which h as some assign ed relation s to these,h as to be foun d ; the quan tity which h as to be foun d iscalled the un kn own quan tity . The relation s are usuallyexpressed in ordinary language in the en un ciation of th eproblem

,and the method o f solving the problem may be

thus described in gen eral terms : den ote the un kn ownqua n tity by th e letter x , an d exp r ess in algebr a ica l

language th e r ela tion s wh ich h old between th e u nkn ow n

quan tity an d th e given qua n tities ; an equa tion will th usbe obtain edf r om wh ich th e value of th e unknown qua n titymay bef o und.

183 . Th e sum o f two numbers is 85 , and their difference is 27 find the numbers.

Let x den ote the less number ; then , sin ce the difference of the numbers is 27, the greater number will beden oted by x + 27 ; and sin ce the sum o f the numbers is 85we have

that is, 2x + 27= 8 5 ;

therefore 2x 85 27 5 8

5therefore2

8

Thus the less number is 29 a nd the greater number is29 27, that is 5 6.

184 . D ivide £ 2. 10s. among A , B , and C, so that Bmay have 5 s. more than A

,and C may have as much as A

and B together.Let x den ote the number o f shillings in A

’s share

,

then x + 5 will den ote the number o f shillings in B ’

s shar e,and 2x + 5 will den ote the number o f shillings in 0 ’

s share.

PROBLEM S . 113

The whole number of shill ings is 5 0 therefore

that is, 4x + 10= 5 0 ;

therefore 4x 5 0 10 40

therefore x 10.

Thus A’s share is 10 shillings, B

'

s share IS 15 shil lings,and 0

’

s share is 25 shill ings.

185 . A certain sum o f mon ey was divided’

between

A ,B

,and C ; A and B to gether received £ 1 7. 1 5 s . A

and 0 together received 1 1 5 s B an d 0 togetherreceived £ 12. 1 0s. find the sumreceived by each .

Let x den ote the number of pounds which A received,then B received 1 7§

—x poun ds because A an d B

together received 1 74 pounds , an d 0 received 1 5 2— 10

poun ds , because A an dp0 together received 1 5 § p ounds.

A lso B an d 0 together received 1 22 pounds ; therefore

2 x + 1 5 i

that is 3 32 29 :

therefore

Thus A received £ 10. B received £ 7. 5 s .,and 0

received £ 5 . 5 3 .

186 . A grocer h as some tea worth 23 . a lh . , and someworth 3s. 6d. a lb. : how many lbs. must he take of eachsort to produce 1001bs . o f a mixture worth 2s. 6d . a lb . 2

Let x den ote the number o f lbs. of the fir st sort ; then100—x will den ote the n umber o f lbs. of the second sort.The val ue o f the x lbs. is 2x shillings and the value of th e

1 14 PROBLEM S .

100 x lbs. is (100—x) shillings. And the whole value

is to be 2x 100 shill ings ; therefore

100= 2x —x) ;

multiply by 2, thus 5 00 4x 700 7x ;

therefore 7x 4x 700 5 00

that is,

3x 200

therefore

Thus there must be 66§ lbs. o f the first sort,and

3 311—1bs . o f the second sort.

187. A lin e is 2 feet 4 in ches long ; it is required todivide it in to two parts, such that on e part may be threefourths o f the other part.Let x den ote the number o f in ches in the larger

‘part ;then 3; will den ote the number o f in ches in the other part .

Th e number o f in ches in the whole lin e is 28 ; therefore

therefore 4x 3x 1 12

x = 16

Thus on e part is 16 in ches long, and the other par t 12in ches long.

188. A person h ad £ 1000, part o f which he len t at4 per cent ., and the rest at 5 per cen t. the whole annual

in terest received was £ 44 : how much was len t at 4 percen t. 1

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1 16 EXAMPLES . XX I.

so lute ly unin telligible, but appear to be susceptible o fmorethan o n e meaning. Th e studen t shoul d then select o n e

mean ing, express that mean ing in Al gebraical symbols, anddeduce from it the result to

o

wh ich it will lead . If t h e

resul t be in adm issible, o r absurd,the studen t should try

an other mean ing o f the words. But if the result is satisfactory he may in fer that he h as probably un derstood thewo rds correc tly ; though it may still be in teresting to trythe other possible mean ings, in order to see if the enunciatio n really 1s sus ceptible o f more than o n e mean ing.

1 91 . A studen t in solving the problems which ar e

given fo r exercise,may find some which he can readily solveby Arithmetic, o r by a process o f guess and trial ; and hemay be thus in clin ed to un dervalue the power o f Algebra,and look o n its aid as un n ecessary. But we may remarkthat by Algebra the studen t is en abled to solve a ll theseproblems, without any un certain ty an d moreover,—h e willfind as he proceeds

,that by A lgebra he can solve pro

blems which would be extremely' difficult o r al togetherimpracticable, if he relied o n Arithmetic alone.

EXAMPLES. XXI.

1 . Fin d the number which exceeds its fifth part by 24 .

+ 1

2. A father 1s 30 years o ld, and)his;son 13

72 years o ld :

m h ow many years will the father be eight timesyas o ld as

the son ?Bo yxand their sum

X 4 . Th e sum o f was raised by A ,B

,and C to ge

ther ; B con tributed £ 15 more than A ,an d 0 £ 20 more

than B : how much did each con tribute Al p - N

(54 S

5 . The differen ce o f two numbers '

1s 14,and their sum

is 4 8 : find the numbers. K “1 1 n6 . A 13 twice as o ld as B

,and seven years ago their

Tun ited ages amoun ted to as man y years as n ow representthe age o f A . find the ag es o f A and B .

x 7 fi x 1 1 5 4

EXAMPLES . XX I.

V 7 . If 5 6 be added to a certain number, the result istreble that number : find the

+ l' t 2 3 X

8 . A child is born in November, and o n the ten th dayo f D ecember he is as many days o ld as the month was on

the day o f h is birth : when was he born ?3 0 ~ X 2 44(1)

9. Find that numberj h ex dpuble o f which in creased byy

’

24 exceeds 80 as much ’

as the number itself is below 100.

A

10. There is a certain fish ’

,tfiéfirea v

ofbwdi ic 1s 9

y‘ in ch es long ; the tail is as long as the head and half theback ; an d the back is as long as the head and tail together : what is the length o f the back and o f the tail} K

1 1 . D ivide the number 84 in to two9fitucit3 § 16h that

if three times o ne part may be equal to four times the other.

3 a:

12 . The sum o f £ 76 was raised by ATB ,zindflbge

ther ; B con tributed as much as A and £ 10 more, an d Gas much as A and B together : h ow much did each con

tribute ?1 3 . D ig

-

e the number 60 in to two parts such that aa seven th o f on e part may be equal to . an eighth o f the otherpart. 7 x 2 (60 k ) y

14 . After 3 4 gallon s had been drawn out of on e o ftwo equal casks , and 80 gallon s o ut o f the other

,there

remain ed just three times as much in on e cask as in theother : what did each cask

1 5 . D ivide the number3 times the greater may exceg 1 6 . A per sOn distributeson s, givin g Sixpen ce each to some, and sixteen pen ce each

y1 7. D ivide the number

the sum o f three times o n e part, and five times the other

part, may be 84 . 3 I1 ~ 8“

k t s—(L18. The price o f a work which comes out in parts is

£ 2. l 6s. 8d. ; but if the price of each part were 1 3 pen cemore than it is, the price o f the work woul be£3 . 7s. 6d. :

11. Q 25 51 010ow many parts were there K” ( fl

19 . D ivide 4 5 in to two parts such that the first dividedby 2 shall be equal to the second mul tiplied by 2.

M a

ri

‘

t fiw - Y )

EXAMPLES . XX I .

20. A father is three times as o ld as h is son ; fouryears ago the father was four times as o ld as his son thenwas : what is the age o f each ?X 4 L1

<z_

V;

N . )(o21 . D ivide 188 in to two parts such that the fourth o f

H i mon e part may exceed the eighth o f the other by 14 .

-

Th ?22. A person meeting a compan y o f beggars gave four

fl pen ce to each, an d h ad s ixteen pen ce left ; he found thathe should have required a shill ing more to en able h im togive the beggars Sixpen ce each : how many beggars werethere ? b e

23 . D ivide 100 in to two parts such that if a third ofon e part be subtracted froma fourth o f the other the r emainder may be 1 1 .

24

5 + 2; 1411 4 2)24 . Two person s

,A an B

, engage at play ; A h as

£ 72 and B h as £ 5 2 when they begin , and after a ce it ainnumber o f games have been wo n and lost between them,

A has three times as much mon ey as B how much did AWin ? 7 1. —mt : 3 \ S

c °

z. X }

25 . D ivide 60 in to two parts such that the differen cebetween the greater an d 64 may be equal to twice thedifferen ce between the less and 3 8 . £11 4 y 2 l U? r W

26 . The sum o f £ 276 was raised by A ,B , and 0 toge

ther ; B con tributed twice as much as A and £ 12 more ,

and 0 three times as much as B and £ 12 more : how muchdid each con tribute l x4 ’

1 X 1—1 1 H K4 0 b f "

”

U 7 A

27. Find a number such that the sum o f its fifth andits seven th shall exceed the sum o f its eighth an d itstwelfth by 1 13 . 1 +

3

7 1 4

28 . An army in a defeat oses o n e - sixth o f its numberkilled and wounded, and 4 000 prison ers it is rein forced

by 3000 men , but retreats, losing o n e - fourth o f its numberin doin g so ; t ere remai n 8000 men

ywh aw the o ri

gin al force ?1<( Xi W O 74 3 m29. Find a number such that the sum o f its fifth and

its seven th shall exceed the differen ce o f its fourth and itsseven th by 994

15 7 6 257m i x A 7

4

7?3 0. On e - half o f a cer tsi in riumber o f person s received

eighteen - pen ce each,o n e - third received two shillings each ,

and the rest received half a crown each ; the Whole sumdistributed was £ 2. 4 s. how many person s were there ?

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EXAMPLES . XX I.

4 1 . In a mixture o f win e an d water the Win e compo sedX 25 gallon s more than half o f the mixture, and the water

5 gallons less than a. third o f the mix ture . how many gallon s were there of each Cl

42. In a.

lo tte1y con sisting o f 10000 tickets, half th eX number of prizes added to o n e - third the number o f blanks

was 3 5 00 : h owmany prizes were there in the lottery fZ / m4 3 . In a certain weight o f gunpowder the

composed 6 lbs. mei e than a half o f the weight, the sulphur5 lbs. less than a third an d the charcoal 3 l . less than a

fourth . how many lbs. were there of each o f the thr eeing1 edien ts ? 4 f m 3 _ 3

4 4 . A gen eral , after h aving lost a‘

battle,found that

he had left fit for action 3 600 men more than half o f hisarmy ; 600 men more than o n e - eighth o f his army werewounded ; and the remainder

,form ing

.

o n e - fifth o f thearmy, were slain , taken prison e1 s, o r missing . what wasthe n umber of the al my .

7

4 5 . How many sheep must a person buy at £ 7 eachthat after paying o ne shilling a score fo r folding them at

n ight he .may gain £ 79 . l 6s. by selling them at £ 8 each i4 6 . A certain sum o f mon ey was shared among five

Kperson s A ,B

, C, D ,and E ; B received £ 10 less than A

0 received £ 16 more than B ; D received £ 5 less than C ;and E received £ 15 more than D ; and it was found thatE received as much as A and B together : how much dideach receive ? 1

‘

L 4 14 7. A tradesman starts with a certain sum o f money

at the end o f the first year he h ad doubled his originalstock, al l but £ 100 also at the end o f the second year hehad doubled the stock at the beginn ing of the second year,all but £ 100 ; also in like mann er at the end o f the thirdyear ; and at the end of the third year he was h ree timesas rich as at first . find h is original stock.

4 8. A person wen t to a tavern wi th a certain sum o f

mon ey there he bo r i owed as much as he h ad about him,

and spent a shilling out of the Whole : with the remainderhe wen t to a second tavern

,where he borrowed as much as

he h ad left,and als o spen t a shilling and he then wen t to

a third tavern , borrowing and spending as before, afterwhich he hadW eft : h owmuch had he at first 3

Q X “ 3 4 N Y 0

PROBLEMS. 121

XXII. Pr oblems, con tinu ed.

1 92. We shall n ow give some examples in which th eprocess of tran slation from ordinary language to algebraical language is rather more difficult than in the exampleso f the preceding Chapter.

193 . It is required to divide the number 80 in to foursuch arts, that the first in creased by 3 , the se cond dimin ish edby 3 , the third multiplied by 3 , and the four thdivided by 3 may all be equal .

Let th e number x den ote the first part ; then if it beincreased by 3 we obtain x + 3 , and this is to be equal tothe second part dimin ished by 3 , so that the second partmust be x + 6 ; again , x + 3 is to be equal to the third partmultiplied by 3 , so that the third part must be

x

;3

x + 3 is to be equal to the fourth part divided by 3 , so thatthe fourth part must be 3 (x + And the sum o f the partsis to be equal to 80.

and

Therefore x x+ 6 3 (x 80,

that is,

2x + 6 +

that is, — 80

mul tiply by 3 ; thus 1 5 x x 3 : 195 ,

that is,

1 6x = 192

therefore

Thus the parts are 12, 18 , 5 ,

122 PROBLEM S .

1 94 . A alon e can perform a piece o f work in 9 days,and B alon e can perform it in 12 days : in what time willthey perform it if they work together ?

Let x den ote the required number of days. In o n e day

A can perform 3th o f the work ; therefore in x days he can

perform gth s of the work. In on e day B can perform

th o f the work ; therefore in x days he can perform

2tbs o f the work . And since in x days A and B to

gether perform the wh ole work, the sum o f the f r actionso f the work must be equal to un ity ; that is,

Multiply by 3 6 thus 4x 3x 3 6 ,

7x 3 6

therefore

195 A c istern could be filled with water by mean s ofo n e pipe alon e in 6 hours, and by mean s o f another pipealon e 111 8 hours ; and it could be emptied by a tap in 12

hours if the two pipes were closed : in what time will thecistern be fill ed if the pipes and the tap ar e all open ?

Let x den ote the required number of hours. In o n e

hour the first pipe fills(

13th of the cistern ; therefore in x

hours it fillsgths of the cistern . In one hour the second

pipe fillséth o f the cistern ; therefore in x hours it fills

8tbs o f the cistern . In o n e hour the tap empties -

1

1—2th

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124 PROBLEM S .

198. A starts from a certain plac e, and travels at therate o f 7 miles in 5 hours ; B starts from the same place8 hours afte r A ,

and travels in the same direction at therate o f 5 miles in 3 hours : h ow far will A travel befo re h eis overtaken by B 2

Let x represen t the number o f hours which A travelsbefo re he is overtaken ; therefore B travels x —8 hours.

Now s ince A travels 7 miles in 5 hours, he travels Z5

mile in o ne hour ; and therefore in x hours he travels

miles. Similarly B travels o f a mile in on e hour, and

5therefore in x—8 hour s he travels53 (

x—8) mil es . And

when B overtakes A they have travelled the same number o f miles. Therefore

multiply by 15 ; thus 25 (x

that is,

25x 200 21x ;

therefo re 25 x 2 1x 200,

that is, 4x 200

therefore

73 : 7Therefore

3 5x 5 0 70 so that A travelled 70 mi les

before h e was overtaken .

199. Problems are sometimes given which suppose th estuden t to have obtained from Arithmetic a kn owledge o f

PROBLEM S . 125

the meaning of p r op or tion ; this will be illustrated in then ext two problems. After them we shall con clude theChapte r with three problems o f a more difficul t characterthan those hitherto given.

200. It is required to divide the number 5 6 in to twoparts such that on e may be to the other as 3 to 4 .

Let the number x den ote the first par t ; then th e otherpart must be 5 6 —x ; and s in ce x is to be to 5 6 — x as 3 to 4

we have

Clear o f fraction s ; thus4x = 3 (5 6—x ) ;

4x : 1 68 — 323 ;

therefore 7x 1 68

therefore

Thus the first part is 24 and the other part is 5 6 24,

that is 3 2.

The preceding method o f solution is the most n aturalfor a beginn er ; the following however is much shorter.Let the number 3x den ote the first part ; then the

second part mus t be 4x ,because the first part is to the

secon d as 3 to 4 . Then the sum of the two parts is equalto 5 6 ; thus

3x + 4x = 5 6 ,

7x = 5 6

therefore x z 8.

Thus the first part is 3 x 8, that is 24 ; and the secondpart is 4 x 8 , that is 3 2.

PROBLEMS.

201 . A cask, A ,con tain s 12 gallons o f win e and 18

gallon s o f water ; and an other cask,B

, con tains 9 gallon so f win e and 3 gall on s o f water : how many gallon s must bedrawn from each cask so as to produce by their mixture7 gallon s o f win e and 7 gallon s of water ?Let x den ote the number o f gallons to be drawn from

A ; then sin ce the mixture is to con sist o f 14 gall ons ,1 4—x will den ote the number o f gallons to be drawn fromB . Now the number of gallons in A is 30, o f which 12 arewin e ; that is, the win e is

c

é—go f the whole. Therefore the

x gallon s drawn from A con tain gallon s o f win e.

Similarly the 14—x gall on s drawn fr omB con tain 9 (1i1

2

— x )

gal lon s o f win e. And the mixture is to contain 7 gallonso f win e ; therefore

12x 9 (14 x )3 0 1 2

that is,

therefore 8x 1 5 (14 x ) 140,

that is,

therefore 7x 70

therefore x 10.

Thus 10 gallons must be drawn from A ,and 4 from B .

202. At what time between 2 o’clock and 3 o’clock isone hand o f a watch exactly over the other ?Let x den ote th e required number o f minutes after

2 o’clock. In x minutes the long hand will m o ve overx division s o f the watch face ; and as the long hand movestwelve times as fast as the short han d, the short hand willmove over x

divisions in x minutes. At 2 o’clock the

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EXAMPLES . XX II:

204 . Four gamesters, A ,B

,O,D

,each with a differen t

sto ck o f mon ey, sit down to play ; A wins half o f B ’s first

stock, B win s a third part o f 0’

s , 0 win s a fourth part o fD

’s, and D win s a fifth par t o f A ’

s ; and then each o f thegamesters has £ 23 . Find the sto ck of each at first.Let x den ote the number o f poun ds which D wo n from

A ; then 5 x will den ote the number in A ’

s first stock.

Thus 4x , together with what A won from B , make up 23 ;therefore 23 — 4 x den otes the number o f pounds which Awon from B . And, sin ce A won half o f B ’

s stock, 23—4xalso den otes what was left with B after his loss to A .

Again , 23 — 4x , together with what B won from 0,make up 23 ; therefore 4x den otes the number of poundswhich B won from C. And

,sin ce B won a third o f 0

’

s

first stock, 12x den otes C ’

s first stock ; and therefore 8xden otes what was left with C after h is loss to B .

Again , 8x , together with what 0 won from D ,make up

23 ; therefore 23—8x den otes the number o f poun ds which0 wo n from D . And, sin ce 0 won a four th o f D

’

s firststock

,4 (23 — 8x ) den otes D

’

s first stock ; and therefore—8x ) den otes what was left with D after h is los s toFin ally, 3 (23 8x), together with x , which D won from

A ,make up 23 ; thus

—8x) +x

23x = 4 6 ;

Thus the stocks at first were 10, 3 0, 24 , 28 .

EXAMPLES. XXII.

1 . A privateer runn ing at the rate o f 10 miles an hourdiscovers a ship 1 8 miles o ff, run n ing at the rate of 8 milesan hour : how many miles can the ship rim before it isover taken l yb

'? “x i—13

55 4: X f / Y

z 7 1.1

2.

“

D ivide the number 5 0 in to two parts such that ifthree - fourths o f o n e part be added to fives ixths

o ther part the sum inlay be 403 7 l f, ( 1 1 )

X 3 . Suppose the distance between London and Edinburgh is 3 60 miles, and that on e traveller starts fromEdinburgh and travels at the rate o f 10 miles an houi ,while an other starts at the same time from Londdn and

travels at the rate o f 8 miles an hour it_

is required to “

know where they will meet. “

3 6 0

M4 Find two numbers 43026?xfige/

rence is 4, and thedifference of their squares4 12

V “ 210 w - r

5 . A sum of 24 sh illinggl yis)received from people ;

some con tribute 9d. each, and some 13 1 d. eachcontributp

r s,were

9th’ere _

éf each kin d ‘22

73.

13“ J

”

.

A” 6. D ivide theQnumber 28 intovtwoz

pafrts such that theexcess of on e part over 20 may be h r ee times th e excess

Y

o f 20 over the other part.(f 1

‘o’

7. A person has £ 98 ; part o f it he lent“at the rate of

5 per cen t. simple in terest,and the rest at the rate o f

6 per cen t. simple in terest ; and the in terest of the wholein 1 5 years amoun ted to £ 81 z h ow3much was lent at 5per cen t i

k; w‘ lI

-

‘

s$5 1 11 " { ii

‘

(A/ o

r 8 . A person len t ifW ain sum o f mon ey at 6 per cen t.simple in terest ; in 10 years the in terest amoun ted to Mless than the sum len t : what w the sum len t 11V fu ‘

w -M . IO V 0 5 K9 . A person ren ts 25 acrgs o f

x

landl for 2701 .2é; theand consists o f two sorts

,the better sort he ren ts at 88 .

per acre, and the worse at 5 3 . per acre h owmany acres ares ! u * ’ M / 3th ere o f each so rt .

“ A s k“ , a ) 12g AJr 10. A cistern tould be filled in 12 minutes by two

ipes which run in to it ; and i t would be filled in 20minutes

" t s

on e could '

,t be filled b the other312113

yd ‘ gn x s

/

1" in to four parts such that/the

(iirs

tflinc

fie

ais

id yby 2, the second dimin ished by 2

,the

thir m ti e y 2, an t e fourth divided b 2 ay 11 1

be equaL W scr - l P’

X Z -i Y -

“y'

_

Ip ai l”

XiS .

LJ 0

12. A person bought 3 0 lbs o f o f two differentsorts, and paid for the whole 19s. better sort cost10d . per lb., and the wo 1 se 7d. per lb.

were there o f each sort 2M 7 :

ga v v 7Tm A/ u y

n vMl O V - l l f o a- V xl

fw e

’

; ”g eu x - f?”

130 EXAMPLES . XX II.

1 3 . D ivide the number 88 in to four parts such thatthe first in creased by 2, the second dimin ished by 3 , thethird multiplied by 4 , and the fourth divided by 5 , may all

7 2be equal ‘ I f KG“; 49

5335Q2 ) g—D

X14 . If 20 men , 4 0 women , and 0 children receive £ 5 0

among them for a week’s work

, and 2 men receive a s muchas 3 women or 5 ch ildren , what does each woman receivefo r a week’

s work ?K‘

1 5 . D ivide 100 in to two parts such that the differenceo f their squares may be 1000.

”L

: XE‘ M D

x 1 6 . There ar e two plac es 5 4 m iles apart, from whichtwo persons start at the same time with a design to meet ;on e travels at the rate o f 3 miles in two hours, and theother at the rate o f 5 miles in four hours when will they

3

5G 17. D ivzide 4 4 iii% two

—”

parts such that the greater ii1creased by 5 may be to the less increased by 7, as 4 is

>\to 3 : 3 x 4 ’ fi

’

fl 7 5 —z b v q —1 9 '

18 . A can do half as much work as B , B can do halfas much as O

', and together they can complete a piece o f

work in 24 days : in what time could each alon e completethe wo rk l ll

f

l

ir t i

L.

4 év ’

;

_ L

19 . D ivide the n'

ul‘

nber 90 in to parts such that ifthe first be in creased by 5 , the second dim in ished by 4 , thethird multiplied by 3 , an d the fourth divided by 2, theresults shall all be equal . 90 a ;

+ 1

20; Three person s can toge thgr complete a piece ofwork in 60 days and it is found that the first does threefourths o f what the second does, and the second four- fifth so f what the third does : in what time co d each on e alon ecomplete the work "

.l

4} 4 , ,3£,

K4 { Q 6

A5 21 . D ivide the number 3 6 in to two parts such that onepart may be five - seven ths of the other . 3 g“ x a

A

K 22. A gen eral o n attempting to draw up h is army 111

the form o f a solid square finds that he has 60 men over,and that he would require 4 1 men more in h is arm in

order to in crease the side o f the square by o ne man : ow

many men were there in the army‘

lh o1 .

{Es 61 >1 ( c +

I/j

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EXAMPLES . XXII.

3 3 . A bill o f £ 3 . 13 . 6d. was paid in half- cro wns, and

and the whole number o f co ins was 28 : how man;ere there o f each kin d Q—{ K f LQL g~ V) g, 1 _ fl

A grocer with 5 6 lbs . o f fin e tea at a lb. woulcoarser sort at 3 3 . 6d. a lh .

,so as to sell the whole

er at 4 3 . 6d. a lb. : what o f the latte r sorttake l lz

tfi z zfl o l

‘

L SP'L

A person hired a labourer to do a certain workagreemen t that for every day he worked he should

but that fo r every day he was absen t he shouldhe worked twice as many days as he was absen t

,

the whole received £ 1. find how many days

36 . A regime:

me time after

295 I

0Wu

J} ; X

X

EXAIIIPLES . XXII .

4 1 . A draper bought a piece o f cloth a'

y ard. He sold on e - third o f it at 43 . per yard,

it at 33 . 8d . per yard, an d th e remainder a

yard ; an d his gai n on the whole was 14 3 . 2d.

y ards did the piece con tain D ig f 8

;4 2. A grazier spen t £ 33 . 73 . 6d . in buy

differen t sorts. Fo r the first sort, which fornof the whole, he paid 93 . 6d . each . Fo r thewhich formed on e - four th of the whole

,he p .

For the rest he}1a1d>123 6d. each

) ;What n

Sun

- adid he buy l f i 3 ) 41 /L

u) _ i qz

4 3 . A market woman boug ht a certain 111

at the ra te o f 5 fo r twopence ; sh e sold half2 a penny , and half of them at 3 a penny, an

by so deing : what wa s the number o f eggs 1q

44 . A pudding cons ists o f 2m o f fiouraisin s

,an d 4 parts of suet ; 7 a lb.

s uet 8d. Find th 11 11he pudding, when the

1 32 EXAMPLES . XXII.

t 3 3 . A bill o f £ 3 . l s. 6d. was paid in half- crowns, and

‘

flo r ins, and the whole number of coin s was 28 : how manycoin s were there o f each kind n i x f LQLS ~ y ) g

3 4. A grocer with 5 6 lbs. o f fin e tea'

at 5 3 . a lb. woulmix a coarser sort at 3 3 . 6d. a lh .

, so as to sell the wholetogether at 4 3 . 6d. a lb. : what uan tity of the latter sortmust he take ?l

_c

fi t s/ 0 i ii: S

fL

A person hired a labourer to do a certain workagreemen t that for every day he worked he should

receive but that for every day he was absen t he shouldlose 9d . ; he worked twice as many days as he was absen t

,

and o n the whole received £ 1. find h ow many daysZ L‘. 6 Q,»

regimen t was drawn up in a solid square ; whenafter it was again drawn up in a solid square

1 7 Bit was found that there were 5 men fewer in a side ; in thein terval 295 men had been removed from the field : what L

was the origin al number o fmen in the r egimen t Cl/ kimy fu 9 0 '

sum o f money was divided between A and B ,

share o f A was to that o f B as 5 to 3 ; al so theo f A exceeded five - n in ths o f the wh ole sum by £ 5 0 :was the share o f each person ? Y z

7 x

A gen tleman left h is whole estate amongson s. Th e share o f the eldest was £ 800 less than hal f ofthe estate ; the share o f the second was £ 120 more thano ne - fourth o f the estate ; the third h ad half as much as

the eldest ; and the youngest had two - thirds o f what thesecond had. Howmuch did each son e 5

3 ?a” 4 1. /

J;2 1

3 9. A and B bggzfii2214

14173?together wit equa sums

mon ey ; A first wo n £ 20, but afterwards lost half o f all.h e then had, and then h is money was halfas much as thato f B : what mon ey had each at first “ f

l bf vf/O)

4 0. A lady gave a giiinba in charity among a numbero f poor, con sisting o f men , women , and children ; each ma n

had 12d, each woman 6d.

,and each child 3d . The number

o f women was two less than twice the number of men ; an dthe number o f children four less than three times thenumber o f women . How many persons were there r e

lieved?/2® a d a a a xyx aL“ 4 0 ) 1 =7—3 ’

L

7 4 1 1 4- 3 1

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13 4

[r l—1. M

4 9 . The squire o f a parish bequeaths a sum equal too n e - hundredth part o f h is estate towards the resto rationKo f the church ; £ 200 less than this towards the endowmen t o f the school ; and £ 200 less than th is latte r sumtowards the Coun ty Ho spitaL After deducting these lega»

g? o f th e estate remain to the heir. What was the4 0

value o f the estate ? 5 L _ 5 040 l Zé1M 4 0

tf ‘

5 0. How many minutes does it want to 4 o’clock, ifthree—quarters of an hour ago it was twice as many minute spast two o’clock ? 7 5 4 r x

‘

z, y

5 1 . Two casks, A and B ,

ar e filled with two kinds o f

1sherry, m ixed in the cask A in the proportion o f 2 to 7,

7

37 I and in the cask B in the propo rtion o f 2 to 5 : what quan

A

2tity must be taken from each to form a mixture whichshall con sist o f 2 gallon s o f the first kind and 6 o f thesecond kind ? ”L ? A

w“

5 2. An officer can form the men o f his regiment in toa hollow square 12 deep. Th e number o f men in thegiment is 1296 . Find the number o f men in the fron t o fthe hollow square . K\ L(

5 3 . A person buys a pieby selling it in allotmen ts finds the value in creased threefold

,so t hat he clears £ 15 0, and retains 25 acres for him‘

self : how many acres were there ?96 2 3 O Y

5 4 . Th e nation al debt of a coun try was in creased'byo fwar . D uring a long peace which

followed £ 25 000000 was paid o ff,and at the end o f that

time the rate o f in terest was reduced from 4 13 to 4 percen t. It was then found that the amoun t of annual inter est was the same as before the war . What was the q .

t o f the debt before gw u-

“ Ji b”

that the loserless than half

quan

tities o f mon ey, Q

rst game andwon the second, he has two shillings more than A : howmuch had each at ‘the commencemen t?

X 435 w 4 /

‘ t J

EXAMPLES . XXII.

5 6 . A clock has two hands turning on the same cen tre ;Athe swifter makes a revolution every twelve hours, and theslower every sixteen hours : in what time will the swiftergain just on e complete revolution o n the slower ? k X

V 5 7. At . what time between 3 o’clock an d 4 o’bl o ck i s

on e hand of a watch exactly in the direction o f the otherhand produced ? Y A .

“ L S‘ i"L

y 5 8. The hands o f a watch are at right angles to eacho ther at 3 o’clock : when are they n ext at right%

i

(1gle

\isC 3

5

0

1

45 9 . A certain sum o f mon ey len t at

’

simplefi

mte est

gamoun ted to £ 297. 123 . in eight mon ths ; and in seven moremon ths it amoun ted to £ 306 : what was the sum ?

60. A watch gain s asmuch as a clock loses ; and 1799hours by the clock ar e equ ivalen t to 1801 hours by the 7watch : find how much the watch gain s and the clock losesW h o“

- L ~£3w3 c

f ?t is between 1 1 and 12 o’clo ck,

Jan 1 is observed

number o f minute spaces between the hands istwo- thirds of what it was ten minutes previously : find thetime. *

( Ai f i b e

2. A and B mjade a j oin t stock o f £ 5 00 by which

gain ed £ 160, o f which A h ad for h is share £ 32 more0

B : what did each con tribute to the stock 1 1 + 06 0

3 . A distiller h as 5 1 gallon s o f Fren ch brandy, whichh im 8 shillings a gallon ; he wishes to buy some Enbrandy at 3 shill ings a gall on to mix with the Fren ch, I 5 1

and sell the whole at 9 shillings a gallon . How many gallon s o f the English must he take, so that he may gain3 0 poi

;

cen t. o n whatah e gave fo r the brandy o f both

f mds 1403 + 3 x 4 7 5 (wow 3 z) : us»

9 f qx1 64 . An officer can form'

his men in to a holl ow squareX4 deep , and also in to a hollow square 8 deep ; the fron t inthe latter formation con tain s 1 6 men fewer than in theformer formation : find the number o f men

136 SIM ULTANEGUS SIMPLE EQUA TIONS .

XXIII. Simul taneous equations of th efir st degr ee wi thtwo unkn own quan tities.

205 . Suppose we have an equation con tain ing two unkn own quan tities x and y , for example 3x—7y = 8 . Fo revery value which we please to assign to on e o f theunkn own quan tities we can determin e the correspondingvalue o f the other ; and thus we can find as man y pairso f values as we please which satisfy the given equation .

Thus, for example, if y z l we fin d 3x = 1 5 , and thereforex = 5 ; if y =2 we find 3x z 22

,and therefore and

so on .

Also,suppose that there is an other equation o f the

same kind, as for example 2x + 5 y z 4 4 ; then we can alsofind asmany pairs o f values as we please which satisfy thisequation .

But suppose we ask for values o f x and y which satisfyboth equation s ; we shall find that there is on ly o n e

'

value

o f x and o n e value o f y . Fo r mul tiply the first equationby 5 ; thus

1 5x—3 5y = 40

and multiply the second equation by 7 ; thus14x 3 5 31 308.

Therefore, by addition ,l 5x - 3 5y

-l 14x + 3 5 y= 40+ 308

that is, 29x : 3 48

therefore

Thus if both equations are to be satisfied x must equal 12.

Put this value o f x in either o f the two given equation s,fo r example in the second thus we obta in

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138 S IM ULTANE O US SIMPLE EQUATIONS .

Then put this value o f a: in either o f the given equations,for example in the second ; thus

108—5 y = 88 ;

20= 5y ;

Suppose,however, that in solving these equations we wish

to begin by el iminating x . If we multiply the first equation by 12 , and the second by 8, we obtain

84y 1200,

40y 704 .

Therefore, by subtr action ,84y 4 0y = 1200 704 ;

124y 496

O r we may render the process more simple ; fo r we mayn

l

iultiply the first equation by 3 , and the seco nd by 2 ;t ius

24x + 2l y = 300,

Therefore, by subtraction ,

21y + 10y = 300 176 ;

3 1y 124

209 . Second method.oExp r ess on e of the unkn ow n

quan tities in ter ms of the o th er f r om eithe r equation ,a nd

substitute th is va lue in th e other equa tion .

Thus,taking the example given in the preceding Arti

cle , we have from the first equationSe = 100 ~ 7y

therefore 100 7y

SIM ULTANE OUS’SIMPLE E QUA TIONS . 1 3 9

Substitute this value o fwin the second equation , and we“

obtain—5 y = 88 ;

that is, 5y 88

3 (100— 7y)—10y = l 76 ;300—21y

3 00

3 l y = 124 ;

y = 4 .

Then substitute this value o f y in either o f the given equation s

,and we shall obtain w= 9 .

O r thus : from the first equation we have7y 100 8x

therefore

Substitute this value o f y in the second equation,andwe obtain

5 (100 8x)7

84x

84x — 5 00+ 40x = 6 16 ;

l 24x = 5 00+ 6 16 : 1 1 16 ;

210. Third method. Exp r ess th e same unkn own

qua n tity in terms of the other f r om each equa tion , andequa te th e exp r essions th us obtain ed .

Thus,taking again the same example

,from the first

100equation a

8and from the second equation

140 SIM ULTANEGUS SIMPLE EQUA TIONS.

100 7g 88 5 y

8 12

Clear o f fractions, by mul tiplying by 24 ;

3 00— 21y z l 76 + 10y ;

3 00—176 : 21y + Ioy ;

3 1y : 124

Therefore

Then,as before, we can deduce x : 9.

O r thus : from the first equation y

from the second equation y

100—8x —88

7 5

From this equation we shall obtain x = 9 ; and then , asbefore, we can deduce y = 4 .

2 1 1 . Solve 19x — 21y = 100, 21x—19y = 14 0.

These equations may be solved by the methods alreadyexplain ed ; we shall use them however to shew that thesemethods may be sometimes abbreviated.

Here, by addition , we obtain19x 21y + 21x

—l 9y z 100 140

40y 240

x —y = 6.

Again , from the original equations, by subtraction , weobtain

21x 19y 1917+ 2l y z 140 100 ;

2y r 4 o

therefore

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EXAMPLES . X X III.

214 . Solve a2a' biy 0

2,

as by 0.

Here a} and y are supposed to den ote unkn own quantities, while the other letters are supposed to den ote kn own

quan tities.

Multiply the second equation by b, and subtract it fromthe first ; thus

that is,

therefore

Substitute

+ by = e ;

c (c—b) e (a —b) —c (c — c).

a —b a —b a—b

c (a—c) c (c—a )g _

b(a — a)'

O r the value of y might be found in the same way asthat o f a: was found.

therefore by c

therefore

EXAMPLES. XXIII.

3x — 4y = 2,

7.y — 5 y = 24 ,

6x — 7y z 4 2,

10x + 9y = 290,

3x—4y = 18,

9 . 4a: 2.v— 3y : 0.

a2x + b?y

— abm—b2y = o2—be ;

a (a b)mz e (e—b) ;e (e b)a (a b)

this value o f a: in the second equation

7x — 9y = 7.

4x—3y = 1 1 .

20x — 3y = 1 .

8x + 9y = 4 1 .

13x l l y = 10.

7x — 6y z 75 .

l l y = 130.

3x + 2y = 0.

14 4

26 3x + 9y z 24,

EXAMPLES XXIII.

'21a'

°

O6y =‘

03 .

27.—6 , 3x—‘

5 y = 28—'25y .

28 .

°

08x

9 4

a; y x y

.r 1 1 4x—5 y

a'+ 1 x — l 6

x_ 1

y— 1 y r

y

y 7x—y 2332. 4x + y _ 1 1

,5 x 1 5

3y—10 (x—1)

+1 0{0 — 5 6

bx—ay z o.

3 5 . bx + ay = 2ab.

a: y_

b+h

"

3 7. (a + c) x—by : bc,

92 ifa+b

"

3 9 . x + y = e, ax—by : c (a—b).

40. a (x 1 .

x —a y—b x + y

—b x—y—a

b a a b

4 2. (a + b)m—(a

(a —b) .fe - l- (a - l- b)y = 2a2—2b2

x y x —y a

'+-

y

a + b a — b_

2ab a2+ b"

'

4 4

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146 SIM ULTANEOUS SIMPLE EQUA TIONS .

We have n ow to find the values o f x and y from (4 )and

Multiply (4) by 9, and multiply (5 ) by 5 ; thus225 x + 171y : 1 13 4

,

225x + 6 5y = 710 ;

therefore,by subtraction ,

106y 424

therefore

Substitute the value o f y in thus76 126

126

Substitute the values o f a: and y in thus14 + 12—2z= 16

thereforetherefore z 5 .

217. Solve 3

x y z

s 4 a .

x y z

Multiply (1) by 2, and add the resul t to (2) thus

a: y z a: y z

SIMULTANE O US SIMPLE EQUA TIONS .1 47

Multiply (1) by 3 , and add the result to (3 ) thus

9 7 8+ 9 _ s+ 14

:e y z .cv y z

that is,10 2:

Multiply (5 ) by 4 , and add the result to thus

4

3?that is

,

therefore 4 7 94x

therefore

Substitute the val ue o f x in thus

therefore 20 17 3(Q

I

therefore

Substitute the values o f x and y in thus

therefore

therefore

148 SIM ULTANEOUS SIMPLE

218. Solve

Subtract (1) from thus

y z x y_

b+c a b

_

that is ,

By subtracting (4 ) from (3 ) we‘

obtain

therefore a:z a .

By adding (4 ) to (3) we obtain

therefore z 3 6 .

By substituting the value of x in (1) we find that y : 2b.

219. In a similar mann er we may proceed if the numb

lero f equation s and unkn own quan tities should exceed

5 rec.

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PROBLEMS .

XXV. Pr oblemswh ich lead to simul tan eous equation sof th efir st degr ee with mor e th an on e unkn own quan tity.

220. We shall n ow solve some problems which lead tosimultaneous equation s of the first degree with more thano n e unkn own quan titv.

Find the fraction which becomes equal to 2when the4

numerator is increased by 2, and equal to 7when the de

n om inator is increased by 4 .

Let reden ote the numerator, and y the den omin ato r ofthe requi red fraction then,by supposition ,

a: 4

y y + 4—7

°

Clear the equations of fractions ; thus we obtain3x — 2y : - 6

Mult iply (1) by 2, and subtract it from thus7x—4 y 16 12

that is,

a: 28.

Substitute the value o f x in thus84 2y 6

therefore 2y= 90 therefore y = 4 5 .

Hence the required fraction is i?221 . A sum of money was divided e ually among a

certain number o f person s ; if there h ad een six more ,each would have received two shillings less than he did ;and if there had been three fewer, each would have r eceived two shillings more than he did : find the number ofpersons, and what each received.

PROBLEM S. 1 5 1

Let x denote the number of persons , and y the numberof shillings which each received. Then xy is the number o fshillings in the sum o f mon ey which is divided ; and, bysupposition ,

— 2) : xy

(x 2)=xy

From (1) we obtainxy + 6y

—2x—12= a'

y

therefore 6y 2x 12

From (2) we obtainxy + 2x

—3y- 6 =xy ;

therefore 2x 3y 6

From (3) and by addition , 3y 18 therefore ySubstitute the value o f y in (4 ) thus

2x 18 6

therefore 24 therefore a' 12.

Thus there were 12 persons, and each received 6

shillings.

222. A certain number o f two digits is equal to fivetimes the sum of its digits , and if n in e be added to thenumber the digits ar e reversed : find the number.Let x den ote the digit in the tens’ plan e, and y the digit

in the un its’ place. Then the number is 10x +y and, bysupposition , the number is equal to five times the sum ofits digits ; therefore

If n in e be added to the number its digits are r eversed,that IS

,we obtain the number l oy + it ; therefore

From (1)we obtain5 x 4y

From(2) we obtain 9x 9 931 therefore a: 1 =y .

PROBLEMS.

Substitute for y in (3 ) thus

5x 4

therefore

Then from (3 ) we obtain y 5 .

Hence the required number is 4 3 .

223 . A railway train after travelling an hour is detain ed24 minutes

,after which it proceeds at six- fifth s o f its

former rate,and arrives 15 m inutes late. If the deten tion

had taken place 5 miles further o n,the train would have

arrived 2 m inutes later than it did . Find the original rateo f the train , and the distance travelled.

Let 5x den ote the number of m iles per hour at whichthe train origin ally travelled, and let y den ote the numbero fmiles in the whole distan ce travelled. Then y

— 5 a' willden ote the number o f miles which remain to he travelledafter the detention . At the original rate o f the train thisdistan ce would be travelled in y

—5 a’

hours ; at the ih

y

the train is detain ed 24 minutes, and yet is on ly 15 minuteslate at its arrival, it follows that the remainder o f thej ourney is performed in 9 minutes less than it would havebeen if the rate had n o t been increased. And 9 minutes

creased rate it will be travelled in hours. Sin ce

69

0o f an hour ; therefore

y—5 x y

— 5 x 9

60

If the deten tion had taken place 5 miles further on ,

there would have been y—5 x—5 miles left to be travelled .

Thus we shall find that

y— 5 x—5 y

—5 x - 5 7

6x 5 x 60o o o o o o o o o o o o o o o o o o o o

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EXAMPLES . XXV.

numbers such that oua is two—thirds o f the other, and theirsum is 100.

We may proceed thus.

“

Let x den ote the greaternumber, and y the less number ; then we have

x +y = 100.

O r we may proceed thus . Let a: denote the greater"

pumber , then 100—x will den ote the less number ; there:ore

3

1 4

O r we may proceed thus. Le t 3x den otenumber

,then will den ote the less number ; therefore

2x 3x 100.

By completing any of these processes we shall find thatthe required numbers are 60 and 4 0.

Th e studen t may accordingly find that he can solvesome o f the examples‘

at the end o f the presen t Chapter,with the aid o f only o n e letter to den ote an unkn own quantity ; and, on the other hand, some o f the examples at theend o f Chapter xxn . may appear to him most naturallysolved with the aid o f two letters. As a gen eral rule itmay be stated that the employmen t o f a larger number o funkn own quan tities renders the work longer, but at thesame time allows the successive steps to be more readilyfollowed ; and thus is more suitable for beginn ers.

The beginn er will find it a good exercise to solve theexample given in Ar t. 204 with the aid o f four letters torepresen t the four unknown quantities which are required.

EXAMPLES. XXV.

1 . If A’s mon ey were in creased by 3 6 shillings he would

have three times as much as B ; and if B ’

s money weredimin ished by 5 shillings he would have half as much as

A : find the sum possessed by each.

2 . Find two numbers such that the first with half thesecond may make 20, and also that the second with a thirdo f the first may make 20.

EXAMPLES . XXV. 1 5 5

3 . If B were to give £ 25 to A they would have equalsums o f mon ey ; if A were to give £ 22 to B the mon eyof B would be double that of A : find the money whicheach actually has.

4 . Find two numbers such that half the first with a

third o f the second may make 32, and that a fourth o f thefirst with a fifth o f the second may make 18.

5 . A person buys 8 lbs. o f tea and 3 lbs. o f sugar fo r£ 1 . 2s. ; and at an other time he buys 5 lbs. o f tea and 4 lbs.of sugar for 1 5 s . 2d. : find the price o f tea and sugar per lb.

6 . Seven years ago A was three times as o ld as B

was ; and seven years hen ce A will be twice as o ld as B

will be : find their presen t ages.7. Find the fraction which becomes equal to when

the numerator is in creased by 1, and equal to i when th e

den ominator is in creased by l .

8 . A certain fishing r o d consists o f two parts ; thelength o f the upper part is to the length o f the lower as5 to 7 an d 9 times the upper part together with 13 timesthe lower part exceed 1 1 times t he whole r od by 36 in ches :find the lengths o f the two parts.

9. A person spends half- a—crown in apples and pears,buying the apples at 4 a pen ny, and the pears at 5 a

penny ; he sell s half his apples and o n e - third o f h is pearsfor 1 3 pen ce

,which was the price at which he bought them :

find how many apples and h ow many pears he bought .10. A win e merchan t has two sorts of win e

,a better

and a worse ; if he mixes them in the proportion o f two

quarts o f the better sort with three o f the worse , themixture will be worth l s. 9d . a quart ; but if he mixes themin the proportion o f seven quarts of the better sort witheight of the worse, the mixture will be worth l s. l od. 3.

quart : find the price o f a quart o f each sor t.

1 1 . A farmer sold to on e person 30 bushels o f wheat,

an d 40 bushels of barley for to another personhe sold 5 0 bushels o f wheat and 30 bushels o f barleyfor £ 17 : find the price o f wheat and barley per bushel.

EXAM'PLES . XXV.

12. A farmer has 28 bushels o f barley at 23 . 4d. a

bushel . with these he wishes to mix rye at 3 3 . a bushel,and wheat at 4 s. a bushel, so that the mixture may con sisto f 100 bushels, and be worth 3 s. 4d . a bushel : find h owmany bushels of rye and wheat he must take.

13 . A and B lay a wager o f 10 shillings ; if A loseshe will have as much as B will then have ; if B loses hewill have half of what A will then have : fin d the mon eyo f each .

14 . If the numerator of a certain fraction be increasedby 1 , and the den ominator be diminished by 1 , the valuewill be 1 ; if the numerator be in creased by the denominator

, and the den ominator dimin ished by the numerato r,the value wil l be 4 : find the fraction .

1 5 . A number o f posts ar e placed at equa l dis tancesin a straight lin e. If to twice the number o f them we addthe distan ce between two con secutive posts, expressed infeet

,the sum is 68 . If from four times the distance be

twe en two consecutive posts,‘

expressed in feet, we subtracthalf the - nun1ber o f posts

,the remainder is 68 . Find the

distan ce between the extreme posts.

16 . A gen tleman distributing mon ey among some poormen found that he wan ted 10 sh illings, in order to beable to give 5 shillings to each man ; therefore he givesto each man 4 shillings on ly, and finds that he h as 5

shillings left : find the number o f po or men an d o f

shillings.

17. A certain company in a tavern found,when they

came to pay their bill, that if ther e had been three moreperson s to pay

the same bill, they would have paid one

shilling each ess than they did ; and if there had beentwo fewer persons they would have paid on e sh illing eachmore than they did : find the number o f person s and thenumber o f shillings each paid.

18 . There is a certain rectangular floor, such thatif it had been two feet broader, and three feet longer, itwould have been sixty- four square feet larger ; but if ithad been three feet broader, and two feet longer, it wouldhave been sixty- eight square feet larger : find the lengthand breadth of the floor.

19 . A certain number of two digi ts is equal to four

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28. A and B run a mile. At the first heat A gives Ba start o f 20 yards, and beats him by 3 0 seconds. At thesecond heat A gives B a start o f 3 2 seconds

,and beats him

by 9fi yards. Find the rate per hour at which A runs.29 . A and B are two towns situated 24 m iles apart,

on the same bank o f a river. A man goes from A to Bin 7 hours, by rowing the first half of the distan ce, andwalking the second half. In return ing he walks the firsthalf at three- fourths of h is former rate

,but the stream

being with h im he rows at double his rate in going ; andhe accomplishes the whole distan ce in 6 hours. Fin d h isrates o f walking and rowing.

3 0. A railway train after travelling an hour is detained1 5 minutes, after which it proceeds at three- fourths o f its

former rate,and arrives 24 minutes late . If the deten tion

h ad taken place 5 miles further o n,the train woul d have

arrived 3 minutes soon er than it did. Find the origin alrate o f the train and the distan ce travelled.

3 1. Th e time which an express train takes to t ravela j ourn ey o f 120 miles is to that taken by an ordinary trainas 9 is to 14 . Th e ordinary train loses as much time insteppages as it would take to travel 20miles without stopping. Th e express train on ly loses half as much time insteppages as the ordin ary train , and it also travels 1 5 mil esan hour quicker. Find the rate o f each train .

3 2. Two train s, 92 feet long and 84 feet long r espec

tive ly, are moving with un iform velocities o n parallel rails ;when they move in opposite directions they ar e observedto pass each other in o ne second and a half; but when theymove in the same direction the faster train is observed topass the other in six seconds : find the rate at which eachtrain moves.

3 3 . A railroad runs from A to O. A goods ’ trainstarts from A at 12 o ’clock, and a passenger train at 1

o ’clock. After going two - thirds o f the distan ce the goods’train breaks down , and can on ly travel at three- fourths o f

its former rate. A t 40 minutes pas t 2 o’clock a co lhs ionoccurs, 10 miles from (J. Th e rate of the passenger trainis double the dimin ished rate o f the goods’ train . Find thedistance from A to C

, and the rates o f the train s.

EXAMPLES . XX V.1 5 9

3 4 . A certain sum o f mon ey was divided between A ,

B,and O’, so that A

’

s share exceeded four- seven ths of theshares o f B and O by £ 30 ; also B ’

s share exceeded threeeighths o f the shares o f A and C by £ 3 0; an d O

’

s shareexceeded two - nin ths o f the shares o f A and B by £ 30.

Find the share o f each person .

3 5 . A and B workin g together can earn 4 0 shillings

in 6 days ; A and 0 together can earn 5 4 shilhngs in 9

days ; and B and 0 together can earn 80 shil lings in 1 5

days : find what each man can earn alon e per day .

3 6. A certain number o f sovereign s, shillings, and six

pen ces amoun t to £ 8. 6s . 6d. Th e amoun t o f the sh ill ingsis a guin ea less than that o f the sovereigns , and a guin eaand a half more than that o f the sixpen ces. Fin d thenumber of each coin .

3 7. A and B can perform a piece of work together in4 8 days ; A and C in 3 0 days ; and B and O’ in 263; days :find the time in which each could perform the work al on e.

3 8. There is a certain number o f th ree digits which isequal to 4 8 times the sum o f its digits, and if 1 98 be subtracted from the number the digits will be reversed ; alsothe sum o f the extreme digits is equal to twice the middl edigit : find the number.

3 9. A man bought 10 bullocks, 120 sheep, and 4 6

lambs. Th e price o f 3 sheep is equal to that o f 5 lambs .

A bullock, a sheep, and a lamb together cost a number ofshill ings greater by 3 00 than the whole number of an imalsbought ; and the whole sum spen t was £ 4 68. 6s . Find theprice o f a bull ock

,a sheep

, and a lamb respectively.

4 0. A farmer sold at a market 100 head o f stock con

sisting o f horses, oxen , and sheep, so that the whole r eah sed£ 2. 73 . per head ; while a horse, an ex

,and a sheep were

sold for £ 22, £ 12. and £ 1 . 10s. respectively. Had hesold on e - fourth the number o f oxen , and 25 more sheepthan he did

,the amoun t received would have been still the

same. Fin d the number o f horses, oxen , and sheep, r espectively which were sold.

QUAD RATIC EQUA TIONS.

XXVI. Quadr a tic Equa tion s.

i226 . A quadratic equation is an equation which co n

tains the squar e o f the unkn own quantity, but no h igherpower.

a] 227. A p ur e quadratic equation is on e which co n tainson ly the square o f the unkn own quan tity. An adf ected

quadratic equation is on e which con tain s the first powero f the unkn own quan tity as well as its square. Thus

, fo rexample, 2w2= 5 0 is a p ur e quadratic equation ; and2x2 7511+ 3 =0 is an adf ected quadratic equation .

228 . Th e following is the Rule for so lving a pquadratic equation . Fin d th e va lue of the squa r e of the

u nkn own quan tity by th e Ru le f o r solving a s imp le equa

tion ; then , by ex tr ac ting th e squar e r oot, the val ues Qf theunkn own quan tity a r ef ound.

x'Z 13 113

2—5

3

Clear o f fractions by multiplying by 30 ; thusl o (fn

2 13) 3 (x2 5 ) 180;

therefore 180+ 130 1 5 z 325

Fo r example, solve

therefore

extract the square root, thus .v : s: 5 .

In this example, we find by the Rule fo r solving a

simple equation ,that x2 is equal to 25 ; therefore a: must

be such a number, that if mul tiph ed in to itself the product is 25 . That is to say, a: must be a square root o f

25 . In Arithmetic 5 is the square root o f 25 ; in Algebrawe ma con sider either 5 o r —5 as a square roo t o f 25 ,since, y the Rule of S ign s

—5 x —5 = 5 x 5 . Hen ce .v

may have either of the values 5 o r —5 , and the equationwill be satisfied. This we den ote thus, .v = 4 5 .

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QUAD RA TIC EQDA TIONS.

230. The following is the Rul e for solving an adfectedquadratic equation . By tr an sp osition and r educ tion

a r r ange th e equa tion so tha t the terms wh ich in volve the

un kn own quan tity ar e alon e on on e side, and the coefi cient

of x2 is + 1 ; add to each side of the equa tion the squa r e

of h alf the coefi cien t Qf x, and then ex tr act the squa r e

r oot of each side.

It will be seen from the examples which we shall n owsolve that the above rule leads us to a poin t from whichwe can immediately obtain the values o f the unkn ownquan tity.

23 1. Solve x2

By transposition , x2 10x

x2—10w+ 5 2=

extract the square root, .v 5 i 1

transpose, mz 5 i l = 5 + l o r 5 - 1 ;

hence .v 6 or 4 .

It is easy to verify that either o f these values satisfiesth e preposed equation ; and it wi ll be useful for the studen t thus to verify h is results.

232. Solve 3x2 4x 5 5 0.

By transpo sition , 3x2 4x 5 5

divide by 3 ,

5 5 4 169

9’

extract the square root, x

tran spose ,

QUAD RATIC EQUA TIONS 163

23 3. Solve 3x 3 5 0.

By tran sposition , 2x2 3x 3 5 ;

2 289.

2 1 6 16’

extract th e square root, .v +

transpose, o r —5 .

234 . Solve .v’ 4.v 1 0.

By transposition , x2 1

x2—4x + 22= 1

extract the square root, .v 2 4 J5 ;

tran spose, x 2 4 J5 .

Here the square root o f 5 cann ot be found exactly ;but we can find by Arithm etic an approximate va1ue o f itto any assign ed degree o f accuracy, and thus obtain thevalues o f .v to any assigned degree o f accuracy.

23 5 . In the examples hitherto solved we have foundtwo d1fi'er en t roots o f a quadratic equation ; in some caseshowever we shall find really on ly on e root. T ake, for example, the equat lon .v

“

by extracting thesquare root we have .v 7 0, therefore .v = 7. It is however found conven ien t in such a case to say that the quadratic equation has two equal r oots.

1 1—2

QUAD RA TIO’ EQUA TIONS .

236 . Solve rez 1 3 0.

By tran sposition 6x 13

add 3 2, 4 .

If we try to extract the square root we have.v 3 i J 4 .

But —4 can have n o square root, exact Or approximate,because any number, whether positive o r n egative, if multiplied by itself, gives a positive result. In this case th equadratic equation h as n o real root ; and this is sometimesexpressed by saying that the roots ar e imagin ary o r

imp ossible.

I 3 123 7. Solve

2 (a'—1)+x2—1 4

.

Here we first clear o f fracti on s by mul tiplying by4 (x

2 which is the least common mult1ple o f the denominators.

- 1 .

By tran sposition , x2—2x = 1 5 ;

add 12, x2

extract the square root, .v 1 4 4 ;

therefore a: 1 4 4 5 o r 3 .

217 +3 23—5 0 12x + 70

1 5 190

Multiply by 5 70, which is the least common multiple o f15 and 190; thus

238 . Solve

3 ( l 2.v +

therefore 210 4 0x ;

therefore 1

’

90(3 .v 5 0) (210 4 0x ) (10 .v )

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QUAD RA TIC _

EQUA TIONS.

240. Ever y quadr atic equa tion can be p u t in th e

form x2+ px+ q

= 0,wh er e p and q r ep r esen t some kn ow n

n umber s, wh ole o r f r action a l, p ositive or n egative.

Fo r a quadratic equation ,by defin ition , con tains n o

power o f the unkn own quan tity higher than the second.

Let all the terms be brought to o n e side, and, if n ecessary,change the sign s o f all the terms so that the coefficien t o fthe square o f the unkn own quan tity may be a po sitivenumber ; then divide every term by this coefficien t

, andthe equation takes the assign ed form .

For example, suppose 7x — 4a 2= 5 . Here we have4x 2 5 O

therefore 4x2 7x 5 0

7x 5therefore4 4

0.

Thus in this example we have p

24 1. SolveBy tran sposition ,

2

2

extract the square root, £23: 4

J(p2

4g)

therefore _ll i

—p * J (p2—4q)

2 2 2

242. We have thus obtained a gener al f ormula forth e roots o f the quadratic equation namely,that .v mus t be equal to

— 4q) or to-

p—J(P

2—4Q)2 2

We shall n ow deduce from this gen eral formula somevery important in feren ces, which will hold fo r any quad

ratio equation , by Art. 240.

QUAD RA TIC ‘

EQUA TIONS . 167

24 3 . A quadr a tic equa tion can n o t have mor e thantwo r oo ts.

Fo r we have seen that the root must be one or theo ther o f two assigned expressions .

244 . In a quadr a tic equation wh er e the terms ar e

a ll on on e side,and th e co efi cien t of th e squar e of th e

un kn own quan tity is un ity , th e sum of th e r oots is equalto th e coefi cien t of th e second term with its sign ch anged,

and th e p r oduct of th e r oo ts is equal to th e last term .

For let the equation be +p .v q 0

the sum o f the roots is—4q)

+

—p—d (p

’—4Q)2 2

that is —p

the product of the roots is—p Jo

e—4g)x

-

p Jun—42)

2 2

PZ—(p

2— 4q)4

that 1s q.

24 5 . The preceding Article deserves special atten tion ,fo r it furn ishes a very good example both of the nature o fthe gen eral results o f Algebra, and o f the methods bywhich these gen eral results are obtain ed. Th e studen tshould verify these results in the case o f the quadraticequations already solved. Take

,for example

,that in

Art . 232 ; the equation may be put in the form

and the roots are 5 and thus the sum o f th e roots is

and the product of the roots is

168 QUAD RA TIC EQUA TIONS .

246 . Solve aw2+ba + 0 : 0.

By tran sposition, av? by

divide by a,

c

+b2 b2—4ac

a 4 a2 4 a2

biN/(b

2—4ac) .

2a

—bi J (b2— 4ao)

2a

extract the square root, .v +

therefore

24 7. The gen eral formulae given in Arts. 24 1 and 24 6may be employed l n solving an

yquadratic equation . Take

for example the equation 3x 4x 5 5 : 0 ; divide by 3 ,thus We have

Take the formula in Ar t. 24 1 , which gives the roots ofand put p :

3,and q=—

5

§-

5; we shall

thus obtain the roots of the proposed equation .

But it 13 more conven ien t to use the formula m Ar t 24 6,

as we thus avoid f1 actions The 131 opo sed equation being3x2 4 x 5 5 : 0

,we must put a : 3

, b z —4,and c : — 5 5

,

in the formula which gives the m ots of + c= 0,

bi (b24 0 6 )that IS, 1n

2a

4 4 J(16 + 660)7

Thuswe have6

that is,

that is, that is, 5 or

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4 9.

EXAMPLES . XXVI .

x + 1 .v ~ 2 9 x + 4+x + 2

.v — l x + 2 5'

.v — 4 .v - 2

(2) — 2 .v—4 14 (Ir—3 x —l

.v — 3 .v—l 1 5°

ctr—2 .v — 4

.v — l .v —3 1 1 1 2 3

.v —4 .v — 2 12° °

.v—2 x + 2 5

3 1 1.v 1 5 — 7.v

2 (.v 2 - 1) 8'

.v 2—1

2x + 1+3x—2 1 1 2x —1 2x

.v — l 3x + 2 2'

.v — 2 6"

3x + 1 2.v - 7 5 2ax—3+32c—5 5

3 (.v - 5 ) 2x—8 2 3 37 —5 2w- 3 2°

3x — 2+2.v - 5 10 x + 2 4—2‘ 7

2x —5 3x—2 3'

.v —l 3

(.v 5 0. —x2).5 1 4 4 5 12

.v + 2 .v x + 4.v + 1 .v + 2 x + 3

x + 1 .v —l —1.v —2 .v + 2

2x + 3

.v + 2 .v —2 .v — l .v + 2 .v —2 .v —3'

.v - 1 5 2 4 5 12d'+ l 6 7 (:c .v + 2 x + 4 .v + 6

.v - 1+

.v —2 2x + 13 x + 1+x + 2 2.v + 13

.v + 1 x + 2 ~.v - l- 16

' '

.v—1 .v—2 x + 1

23;—1 3x —1 5 x—1 1 14 32—9 {02—3x + 1 .v + 2 .

fc—l

60’ x8x — 3

“

a'+- 1

°

a2x2—2a3x + a‘ 62. 4a2x = (a 2

.v a=x+b

.1 +

1 1 +1

a w b a’

EQUATIONS LIKE QUAD RA TICS. 171

XXVII. Equa tions wh ich may be so lved likeQuadr atics.

2 48. There ar e many equation s which are n o t strictlyquadratics, but which may be solved by the method o f comp leting th e squar e ; we will glve two examples.

24 9. Solv‘e x 6 8.

.v 7x3

extract the square root, a s

therefore w3 o r —1 ;

extract the cube root,thus x = 2 o r 1.

25 0. Solve A) ? 3x 3 J(x2+ 3x—2) 6.

Subtract 2 from both sides,thus

x2+ 3x

Thus on the left- hand side we have two expressions,

namely, J(x2+ 3x—2) and a" 3x 2,and the latte r is th e

square o f the former ; we can n ow comp lete the squa r e.

a2+ 3x q (x2+ 3x

extract the square root, thus'

J (a2+ 3x

therefore d a;? 3x 2) 1 o r 4 .

172 EQUA TIONS LIKE QUAD RA TICS.

First suppose J(.v2 1 .

Square both sides, thus x2 3x 2 1 .

This is an ordinary quadratic equation ; by solving it—3 4 .J21

we shall obtain x :

2

Next suppose J(x2 3 .v 2) 4 .

Square both sides, thus x2 3x 2 1 6.

This is an ordinary quadratic equation ; by solving itwe shall obtain .v = 3 o r 6 .

Thus o n the whole we have four values for .v , namely,3 o r —6 0r

2

An important observation mus t be made with respectto these values. Suppose we proceed to verify them .

If we put x = 3 we find that and thusN/(wh 3x $ 4 . If we take the value 4 the originalequation will n o t be satisfied ; if we take the val ue —4 itwill be satisfied. If we put a}

: — 6 we arrive at the sameresult. And the result might have been an ticipated,because the values .fv = 3 or 6 were obtain ed fromN/(x

2-1- 3x —4

, which was deduced from the originalwe find that

x2+ 3x and the original equation will be satisfiedif we take S (x2+ 3x + 1 ; and, as before, the resultmight have been an ticipated.

equation . If we put a:

In fact we shall find that we arrive at the same fourvalues o f .v , by solving either of the foll owing equations,

mg+ 3a—3J (a3 + 3x

but the values 3 o r —6 belong strictly on ly to the first- 3 4 .J2 1

2belong strictly only toequation

,and the values

the second equation .

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174 EQUA TIONS LIKE QUAD RA TICS.

25 5 . Equations are sometimes proposedintended to be solved, partly by in spection , and partly byordinary methods ; we will give two examples.

.v + 4 .v —4 9 + x 9 — .v

25 6“ So lvex _ 4 .v + 4 9—.v 9 + .v

'

Bring the fractions on each side of the equation to acommon den ominator ; thus

(9 + x)’

8 1 -

.v2

16x 3 6xthat 18,332— 16 81—.v

2°

Here it is obvious that x = 0 is a roo t. To find theother roots we begin by dividing both sides of

tion by thus4 9

a“ 1 6 81 x

2

therefore 4 (81 x2) 9 (a:

2 16)

therefore 13x2 3 24 14 4 4 68

therefore .v’ z 3 6

therefore .v 4 6 .

Thus there ar e three roots o f the proposed equation,

namely, 0, 6 , —6 .

25 7. Solve .v’ 7.va

26a3 0.

Here it is obvious that .v = a is a root. We maywrite the equation x

3 —a) and to find th eother roots we begin by dividing by .v — a . Thus

-t

By solving th is quadratic we shall obtain m= 2a o r 30 .

Thus there are three roots o f the proposed equation ,namely, a , 2a , — 3a.

V

EXAMPLES . XXVII . 170

EXAMPLES. XXVII.

1 . 34 - 1 3x24 3 6 = o. 2 . x—5 4 x

3 .4 .

5 . 2J(x’

.v“ 7. J (x

2

9 J(x2

2x2+ 6x = 226'

w4—4x2— 2J (w4

13 . m= 7J (2—w2).

—3 . 15 . J(x + 8)

5 S (I

J (3x

J (2x + 1 ) +J(7x

J (b2+ ax)

2xJ (a .v’

) 2m2 z a2—a .

1 1 3x

—w)'

1—.v 1 + .v 1 + x 2‘

1 1 1 1

x + 7 .v — l x + 1 .v—7

1 1

—.v2) x —J (2—x 2)

2_1 ) .v l )— 8 J(.v

2— l ).x—J(x

2—1 ) —1)

m+ a x—a b+ .v b—x

x —a x + a b—x b+ x'

27. ma 3dmz 4 a3. 28 . 5 x2(a .v ) (a2

292) (.v 3a ).

XXVIII. Pr oblems wh ich lead to Quadr aticE qua tion s.

25 8. Find two numbers such that their sum is 1 5 ,and their product is 5 4 .

Let .v den ote on e o f the numbers, then 15 —.v willden ote the other number ; and by supposition

.v (15 x) 5 4 .

By transposition , x2 5 4

therefore 502 5 4

therefore .v

3therefore -

2

— 9 or 6.

If we take .v : 9 we have 1 5 — .v = 6, and i f we take6 we have 1 5 —

.v = 9 . Thus the two numbers are 6 and 9 .

Here although the quadratic equation gives two values o f

.v , yet there is really on ly o n e solution o f the problem .

25 9. A pe1 son laid out a certain sum of mon ey in

goods, which he sold again for £ 24 , and lost as much pe1cen t. as he laid ou t : find h owmuch he laid out.

Let .v den ote the n umber of pounds which he laid out

then x —24 will den ote the number of poun ds which helost. Now by supposition he lost at the rate o f .v per cen t .,that is the loss was the fraction .v

o f the cost ; therefore

.v 24

therefore "3 2400.

From this quadratic equation we shall obtain a': 4 0

o r 60. Thus all we can in fer 1s that the sum o fmon ey laido ut was either £ 40 o r £ 60 ; for each o f these numbe rssatisfies all the conditions o f the problem.

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PROBLEMS.

262. In solving problems it is often found, as in Art . 260,that results are obtain ed which do n o t apply to the problemactually proposed. The reason appears to be, that thealgebraical mode of expression is more gen eral than ordinary language, and thus the equation which is a preperrepresen tation of the con dition s o f the problem will alsoapply to other cond ition s. Experien ce will convin ce thestuden t that he will always be able to select the resultwhich belongs to the problem he is solving. And it will beoften possible, by suitable changes in the enun ciation o f theo riginal problem , to form a n ew problem corresponding toany result which was in applicable to the original problem ;this is illustrated in Article 26 1, and we will n ow give an other example .

263 . Find the price o f eggs per score, when ten morein half a crown ’

s worth lowers the price threepen ce perscore.Let x den ote the number o f pen ce in the price o f a

a:score of eggs, then each egg costs

20pen ce and therefore

the number of eggs which can be bought for hal f a crowna: that is 62

0If the price were threepence

per score less, each egg would cost W pence, and thenumber o f eggs which could be bought for hal f a crown

"

would be600

”c 3

Therefore,by supposition ,

600 600

therefore 60 (w

therefore x2 3x 180.

From this quadratic equation we shall obtain mz 1 5

o r —12. Hen ce the price required is l sd . per score . Itwill be found that 12d . is the resul t o f the following prablem ; find the price o f eggs per score when ten f ezr erin half a crown ’

s worth r aises the price threepence perSCOY O .

EXAMPLES . XXVIII . 179

EXAMPLES. XXVIII.

1 . D ivide the number 60 in to two parts such thattheir product may be 864 .

2. Th e sum o f_two numbers is 60, and the sum o f

their squares is 1872 : find the numbers.

3 . Th e differen ce o f two numbers is 6,and their pro

duct is 720 : find the numbers.

4 . Find three numbers such that the second shall betwo- thirds of the first

,and the third half of the first ; and

that the sum o f the squares o f the numbers shall be 5 4 9.

5 . Th e differen ce of two numbers is 2, and the sum o f

their squares is 244 : find the numbers.

6 . D ivide the number 10 in to two parts such thattheir product added to the sum o f their squares may make76.

7. Find the number which added to its square rootwill make 210.

8 . On e number is 16 times an other ; and th e productof the numbers is 14 4 : find the numbers.

9 . One hun dred and ten bushels o f coals were dividedamong a certain number of poor person s ; if each per so nhad received on e bushel more he would have received as

many bushels as there were person s : find the numberof persons.

10. A company dining together at an inn find theirbill amoun ts to £ 8 . two o f them were n o t allowed topay, and the rest found that their shares amoun ted to 10shillings a man more than if all had paid : find the

'

number

ofmen in the compan y.

1 1 . A cistern can be supplied with water by twopipes ; by on e of them it would be filled 6 hours soon erthan by the other, an d by both together in 4 hours : findthe time in which each pipe alon e would fill it.

12 -2

PROBLEMS.

262. In solving problems it is often found, as in Art . 260,

that results are obtain ed which do n o t apply to the problemactually proposed. The reason appears to be, that thealgebraical mode o f expression is more gen eral than ordinary language, and thus the equation which is a properrepresen tation of the conditions o f the problem will alsoapply to other cond ition s. Experien ce will convin ce thestuden t that he will always be able to select the resultwhich belongs to the problem he is solving. And it will heoften possible, by suitable changes in the enunciation o f theorigin al problem , to form a n ew problem correspondin g toany result which was in applicable to the original problem ;this is illustrated in Article 26 1, and we wil l n ow give an other example .

263 . Find the price o f eggs per score, when ten morein half a crown ’

s worth lowers the price threepence perscore.Let x den ote the number o f pen ce in the price o f a

score o f eggs, then each egg costs 57) pen ce and thereforethe number of eggs which can be bought for half a crown

m that is 600 If the price were threepen ce

per score less, each egg would cost W pence, and thenumber of eggs which could be bought for hal f a crown

‘

would be 60°

"

0 3Therefore

,by supposition ,

600 600

G

therefore 60 (.i

therefore aL’ 180.

From this quadratic equation we shall obtain 33 : 1 5

or —12. Hen ce the price required is 1 5 d. per score. Itwill be found that 12d . is the result of the following prdblem ; find the price o f eggs per score when ten f ewerin half a crown ’

s worth r aises the price threepence perscore.

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EXAMPLES . XX VIII.

12. A person bought a certain number o f pieces o fcloth for £ 3 3 . which he sold again at £ 2. 83 . pe r piece,and he gain ed asmuch in the whole as a single piece cost :find the number of pieces o f cloth.

13 . A and B together can perform a piece o f work in14%days ; and A alon e can perform it in 12 days lessthan B al one : find the time in which A alon e can pe rform it.

14 . A man bought a certain quan tity o f meat for18 shillings. If meat were to rise in price o n e pennyper lh .,

he would ge t 3 lbs. less for the same sum. Findh owmuch meat he bought.

1 5 . The price o f on e kind of sugar per ston e o f 14 lbs.

is 1 3 . 9d. more than that o f an other kind ; and 8 lbs. less o fthe first kind can be go t for £ 1 than of the second : findthe price o f each kind per ston e.

16 . A person spen t a certain sum o f mon ey in goods,which he sold again for £ 24 , and gain ed as much per cen t.as the goods cost him : find what the goods cost.

17. The side o f a square is 1 10 in ches long : find thelength and breadth o f a rectangle which shall have itsperimeter 4 in ches longer than that o f the square, and itsarea 4 square in ches less than that o f the square.

18 . Find the price of eggs per dozen , when two less ina shilling’

s worth raises the price o n e penny per dozen .

19 . Two messengers A and B were despatched at thesame time to a place at the distance o f 90 miles ; theformer by riding o n e mile per hour more than the latterarrived at the end o f h is j ourn ey o n e hour before him : find

at what rate per hour each travelled.

20. A person rents a certain number o f acres o f pas

ture land for £ 70 he keeps 8 acres in his own possession,

and sublets the remainder at 5 shillings per acre more thanhe gave, and thus he covers h is ren t and has £ 2 eve r :

find the number of acres.

EXAMPLES . XX VIII. 181

21 . From two places at a distan ce o f 3 20 miles, two

person s A and B set out in order to meet each other.A

'

travelled 8 miles a day more than B ; and the number o fdays in which they met was equal to half the number o fmiles B wen t in a day. Find h ow far each travelled beforethey met.

22. A person drew a quan tity o f win e from a full vesselwhich held 8 1 gallons, and then filled up the vessel withwater. He then drew from the mixture as much as bebefore drew of pure win e ; an d it was found that 64 gallonsof pure wine remained. Find how much he drew each time.

23 . A certain company of soldiers can be formed intoa solid square ; a battalion con sisting o f seven such equalcompan ies can be formed into a hollow square, the menbeing four deep. The hollow square formed by the battalion is sixteen times as large as the solid square formedby o n e company. Find the number ofmen in the company.

24 . There ar e three equal vessels A ,B

,and C

'; the

first con tain s water, the second brandy, and the thirdbrandy and water. If the con ten ts o f B and C be puttogether, it is found that the fraction obtain ed by dividingthe quan tity of brandy by the quan tity of water is n in etimes as great as if the con ten ts o f A an d 0 had beentreated in l ike mann er. Find the preportion o f brandy towater in the vessel C.

25 . A person len ds £ 5000 at a certain rate of in terest ;at the end o f a year he receives h is in terest, spends £ 25 o f

it, and adds the remainder to h is capital ; he then lendshis capital at the same rate“

o f in terest as before, and atthe end o f an other year finds that he has altogether£ 5 382 : determin e the rate o f in terest.

SIM ULTANEGUS EQUATIONS

XXIX. Simultan eous Equation s in volving Quadr a tics.

264 . We shall n ow solve some examples o f simultaneous equation s in volving quadratics. There ar e two cas eso f frequen t occurren ce for which .rules can be given ; inboth these cases there ar e two un kn own quan tities and twoequa tion s. The unkn own quan tities will always be denotedby the letters .v and y .

265 . Fir st Case. Suppose that on e o f the equation sis o f the first d egree, and the other o f the second degree .

Rule. Fr om th e equa tion of th e fir st degr eefin d th evalue of eith er of th e unknown quan tities in terms ofth e oth er , an d substitu te th is va lue in th e equa tion ofth e secon d degr ee.

Example . Solve 3x 4y 18, Say z: 2.

18 - 3 .v

4

value in the second equation ; thereforeFrom the first equation y : substitute this

therefore 2owz 9x2 8

therefore 5 4x 8.

From this quadratic equation we find .v = 2 or

then by substituting in the value o f y we find y = 3 or

266 . Solve 3w2+ 5x—8y = 3 6, — 3x — 4y z 3 .

Here although n either of the given equations is o f thefirst degree,

Iyet we can immediately deduce from them an

equation of t e first degree .

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SIM ULTANEO US EQUA TIONS

From this quadratic equation we shall obtain v= 2 o r 3 .

In the equation x2(1 put 2 fo r v thus

a : and since y = vx , we have y = £ 4 . Again, in thesame equation put 3 for v thus a : 4 J2 and sincey = vx

,we have

O r we might proceed thus : mul tiply th e first of thegiven equation s by 2 ; thus

2x2+ 2xy + 4312: 88

the second equation is xy y? 1 (3,

By subtraction 3y2 72, therefore y? 24 .vy .

Again , multiply the second equation by 2 and subtractthe first equation ; thus

3x2 3xy 12 therefore x2 : .

‘

cy 4.

Hence, by mul tiplication£02

212(24 we) (we

2:3 2y2 28x31

By solving this quadratic we obtain .vy = 8 o r 6 . Substitute the former in the given equations ; thus

x2 2y

2 3 6, y

2= 24 .

Hence we can find x2and y

2. Similarly we may take the

other value o fmy , and then find x2 and yg.

268. Solve 2x2+ 3xy y2 z 70, 6402+ xy

—y?= 5 0.

Assume y = vx,and substitute for y ; thus

x2(2 x2 (6 + v

Therefore by division2 + 3v+ v2 70 7

6 + v— v2 5 0 5’

thereforetherefore 12v2 81: 32 0

therefore 3v2 2v 8 z 0.

IN VOLVIN G QUAD RA TICS. 185

From this quadratic equation we shall find v o r - 2 .

In the equation put 2for v ; thus.v : 4 3 ; and since y = v.v we have y : 4 4 . The valuev= - 2 we shall find to be inapplicable for it leads to theinadmissible resul t a}? x 0z 70. In fac t the equations fromwhich the value o f v was obtain ed may be written thus,

and hen ce we see that the value o f v found from 2 +v= 0

is inapplicable, and that we can on ly have

Z and therefore v3 —v 5 0 5

’

269. Equation s may be propo sed which do n o t fallun der either o f the two cases which we have discussed,but which may be solved by ar tifices which can on ly besuggested by trial and experien ce. We will give someexamples.

270. Solve x + y z 5,

x3 +y

3= 65 .

By division ,6

5

5

x2

.vy y2= 13

then from this equation combin ed with x + y = 5 we can

find a and y by the first case. O r we may complete thesolution thus,

x +y z 5

square x2 2xy y

2 25

Also x2—xy + y

2= l 3

Therefore,by subtraction , 3xy 12

therefore .vy = 4 ;

therefore 4xy 16

Subtract (3) fr om (1 ) thus2xy y

2 9

extract the square root,

.v y 4 3 .

186 SIM ULTANEOUS EQUATIONS .

We have now to find .v and y from the simple equations3 : y z i 3

these lead to

271. Solve x2+ y

2= 4 l , .vy = 20.

These equation s can be solved by the second case ; o rthey may be solved in the mann er just exemplified. Fo rwe can deduce from them

a2+ y

2+ 2xy = 4 l

x2+y

z—2xy = 4 1—4 0= 1 ;

then by extracting the square roots,.v—y z 4 1 .

And thus fin ally we shall obtainor 4 4

, y = 4 4 o r =h 5 .

272. Solve a;‘ 2

.'v“4 - l- y

4 133 .

a4+ x2y

2+ y4 13 3

1 9’

that is, wB—xy -Fy'zz l

We have n ow to solve the equationsxZ

By addition and subtraction we obtain successivelyx2+ y

2: 13 , .vy = 6 .

Then proceeding as in Art . 271 , we shall findwe S o r 4 2

, y = =h 2 or 4 3 .

By division ,

273 . Solve x .v5 —y5z 242.

x5—y5 242

.

.v —y 2

x —y = 2 ;

x2—2xy + y2= 4

x2+ y

2=23 y + 4

By division ,

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188 EXAMPLES .

'

.XXIX.

x+J

a+b

_ 2, xy = ab.

2,

d 2+ evy = 28, xy—ygz 3 .

y2+ xy z 3 6 .

2x2—xy z 5 6 , 2xy—y2 r - 4 8.

—2.xy z 1 5 , wy— 2y

2= 7.

.v2+ .vy

— 6y2: 21

,ay

—2y9= 4

.

x2+ 3xy = 5 4 , xy + 4y

2= 1 15 .

x + y x —y 5.v —

y .v + y 2’

. .v2+ 2 25

zz_ z2 — 7 , wy = 48o

x?y2= 90.

w+ y .v—y 10

.v —y .v + y 3

x2y (2.v

2x2 2d 2+ 4xy = 5 y .

x + y .v —y a 2+ 1

’ x2+ y2z b3.

.v—y .v + y a

x2+ y

2: a

2+ bz.

x2+ 2xy y

2= a2+ 2a—1 ,

(a —y)

.v —y = 2, a3 —y3 = 15 2.

$2 _

y2= 3 0

PROBLEMS .

XXX. Pr oblems wh ich lead to Quadr atic Equa tionswith mor e than o n e u nkn own qu an tity .

274 . There is a certain number o f two digits ; the sumo f the squares o f the digits is equal to the n umber ihcreased by the product o f its digits ; and if thirty- six be

added to' the number the digits are reversed : find the

number.Let a: den ote the digit in the ten s’ place, and y the

digit in the un its’ place. Then th e number is 10.v + y ; and

if the digits be reversed we obtain 10y + az Therefore, bysupposition , we have

From (2) we obtain9y = 950+ 3 6 ; therefore y = x + 4 .

Substitute in thus

therefore x2 7x + 12 0.

From this quadratic equation we obtain .v = 3 o r 4 ;and therefore y = 7 o r 8 . Hen ce the required numbermust be either 3 7 o r 4 8 ; each o f these numbers satisfiesal l the condition s o f the problem .

275 . A man starts from the foot o f a moun tain towalk to its summit. His rate o f walking during thesecond half o f the distance is half a mile per hour less thanh is rate during the first half, and he reaches the summit in5 5 hours. He descends in 3 5} hours by walking at a un i

form rate,which is o n e mile per hour more than h is rate

during the first half o f the ascen t. Find the distance tothe summit, and h is rates of walking.

Let 2a: den ote the number o f miles to the summit,a nd

suppose that during the first half o f the ascen t the man

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EXAMPLES . XXX .

EXAMPLES. XXX.

1 . The sum o f the squares o f two numbers is 170, andthe differen ce o f their squares is 72 : find the numbers.

2. Th e product o f two numbers is 108, and their sumis twice their difference : fin d the numbers.

3 . The product o f two numbers is 192, and th e sum o f

their squares is 640 : find the numbers.4 . The product o f two numbers is 128 , and the differ

en ce o f their squares is 192 : find the numbers.

5 . The product o f two numbers is 6 times their sum,

and th e sum of their squares is 325 : find the numbers.

6 . Th e product o f two numbers is 60 times their differen ce, and the sum o f their squares is 244 : find the numbers.

7. Th e sum of two numbers is 6 times their differen ce,and their product exceeds their sum by 23 : find the n umbers.

8 . Find two numbers such that twice the first withthree times the second may make 60, and twice the squareo f the first with three times the square of the second maymake 840.

9 . Find two numbers such that their differen ce multiplied in to the difference o f their squares shall make 32,and their sum mul tiplied in to th e sum o f their squaresshal l make 272 .

10. Find two numbers such that their difference addedto the differen ce o f their squares may make 14 , and theirsum added to the sum o f their squares may make 26.

1 1 . Find two numbers such that their product is equalto their sum, and their sum added to the sum of theirsquares equal to 12.

EXAMPLES . XXX .

12. Find two numbers such that their sum in cr easedby their product is equal to 3 4 ; and the sum o f theirsquares dimin ished by their sum equal to 4 2.

1 3 . Th e difference o f two numbers is 3 , and the difference o f their cubes is 279 : find the numbers.

1 4 . The sum o f two numbers is 20, and the sum o f

their cub‘es is 2240 : find the numbers.

1 5 . A certain rectangle con tain s 3 00 square fee t ; a

second rectangle is 8 feet short er, an d 10 feet broader,and also con tain s 300 square feet : find the length and

breadth of the first rectangle.

16 . A person bought two pieces o f cloth o f differen tsorts ; the fin er cost 4 shill ings a yard more than thecoarser, an d he bought 10 yards more of the coarser thano f the fin er. Fo r the fin er piece he paid £ 18, and for thecoarser piece £ 16 . Find the number o f yards in each piece.

17. A man h as to travel a certain distance ; an d whenhe h as travelled 4 0 miles he increases h is speed 2 m ilesper hour. If he h ad travelled with h is in creased spe edduring the whole o f his journ ey he woul d have arrived 4 0minutes earlier ; but if he h ad con tin ued at his originalspeed he would have arrived 20 minutes later. Find th ewhole distan ce he h ad to trave1, and his original speed.

1 8 . A number con sisting of two digits has on e decimalplace ; the difier en ce o f the squar es of the digits is 20, andif the digits be reversed, the sum o f the two numbers is 1 1 1find the number.

1 9 . A person buys a quan tity of wheat which he sellsso as to gain 5 per cen t. on h is outlay, an d thus clears £ 16~If he h ad sold it at a gain o f 5 shi ll ings per quarter, hewoul d have cleared as many poun ds as each quarter costh im shil lin gs : fin d how many quarters he bought, and

what each quarter cost.

20. d workmen,A and B

,were employed by the

day at differen t rates ; A at the end o f a certain numberbf days received £ 4 . l 6s.

,but B

,who was absen t six o f

T A.

EXAMPLES. XXX.

those days, received on ly £ 2. 14 3 . If B had worked thewhole time

,and A h ad been absen t six days, they would

have received exactly alike . Find the number o f days,and what each was paid per day.

21 . Two train s start at the same time from two towns ,and each proceeds at a uniform rate towards the othertown . When they meet it is found that o n e train h as run108 miles more than the other, and that if they con tinueto run at the same rate they will fin ish the journ ey in 9 and1 6 hours respectively. Find the distance between thetowns and the rates of the train s.

22. A and B are two town s situated 18 miles apart onthe same bank o f a river. A man goes from A to B in4 hours, by rowing the first half of the distan ce and walkingthe second half. In return ing he walks the first hal f atthe same rate as before, but the stream being with h im ,

herows 15 miles per hour more than in going, and aecomp lish es the whole distan ce in 3 5 hours. Find h is rates o f

walking and rowing.

23 . A and B run a race round a two mile course. Inthe first heat B reaches the winning post 2 m inutes beforeA . In the second heat A in creases his speed 2 m iles perhour

,and B dimin ishes h is as much ; an d A then arrives

at the winn in g post two min utes before B . Find at whatrate each man ran in the first heat.

24 . Two travellers, A and B,set out from two places,

P and Q, at the same time ; A starts from P with thedesign to pass through Q, and B starts from Q and travelsin the same direction as A . When A overtook B it wasfound that they h ad together travelled thirty m iles, thatA h ad passed through Q four hours before, and that B ,

at

h is rate o f travelling, was n in e hours’ j ourn ey distan t fromP. Find the distance between P and Q.

“

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IN VOLUTION.

Thus, fo r example,(a2b3 )z a

4b6 ; a2b3)

3a6b9 (ab

zo3)4

a4b801 2 ;

“26309

6a10b1 5c20; (2ab

2c3)6 26611

65 12018 64 61617120180

280. Rule for obtain ing any power o f a fraction . Raise

both th e n umer ator and den om in ator to tha t p ower , andgive th e p r op er sign to th e r esult.

This follows from Ar t. 14 5 . Fo r example,4 24a8 1 6a8

34174 8 1b"

281 . Some examples o f Involution in the case o f

bin omia l exp r ession s have al ready been given . See

Arts. 82 and 88 . Thus2ab+b2,

(a b)3r a

3 3uzb 3ab2 b3.

The studen t may fo r exercise obtain the fourth, fifthand sixth powers o f a + b. It will be found that

(a b)4

a4 4a3b 6a2b2 4ab3 b‘

,

(a b)5

as 5 a4b 10a3b2 10a2b3 5 ab4 b5.

(a b)6

a6 6a

5b 1 5 a 4b2 2Oa3b3 l 5 a’b4 6c 5 b“.

In l ike mann er the following resul ts may be obtain ed :

(a b)2= a

'3 2ab+ b

(a b)3

a3 3a2b 3ab2 b"

,

(a b)4 = a

4 4 a3b Gagbg 4 ab3 b‘,

(a b)5

a 5 a4b 10a3b2 10a5

'b3 5 ab“ bs.

(a - b 6 as Gasb 1 5 a "b2 20a 3b3 Gab5 b6

Thus in the results obtained fo r the powers o f a b,

where any odd power of b occurs,the n egative sign is pre

fixed and thus any power o f a b can be immediatelydeduced from the same power o f a + b, by changing thesign s of the terms which involve the odd powers of b.

INVOLUTION . 197

282. The studen t will see hereafter that, by theQ '

aido f a theorem called the B in omial Theor em,

any powero f a bin omial expression can be obtain ed without th elabour o f actual mul tiplication .

283 . Th e formulae given in Article 28 1 may be usedin the way we have already explain ed in Art. 84 . Supo se, for example, we require the fourth power o f 2x—3y .

n the formula for (a b)4put 2x for a, and 3y for b ; thus,

3y)“(259

4 4 (25 036 9 ) 6 4 (23 96 9?

16x4 96x3y 216x2y2 216xy

3 81y4.

284 . It will be easily seen that we can obtain requiredresults in Involution by differen t processes. Suppose

,fo r

example,that we require the sixth power o f a + b. We

may obtain this by repeated multiplication by a + b. O r

we may first find the cube o f a + b, and then the square o fthis result ; sin ce the square of (a + b)3 is (a + O r wemay

‘

first find the square o f a + b, and then the cube o f thisresult ; since the cube of (a b)

2 is (a In l ike mann erthe eighth power of a + b may be found by taking thesquare o f (a + b)4, o r by taking the fourth power o f

285 . Some examples of Involution in the case o ftr in omia l expr ession s have already been given . SeeA rts. 85 and 88. Thus

These formulae may be used in the mann er explain ed inA rt . 84 . Suppose, for example, we require (1In the formula for put 1 for a,

—2x for b,and

3x2 for 0 ; thus we obtain(1 2x 3x2)?

- 2x)2+ (3az)z 2x) 2 (1 + 4x2+ 9x4—4x—12x3 + 6x2

1 - 4x + 10x2—12x3 + 9x 4.

EXAMPLES . XXX] .

Similarly, we have(1 2x + 3x2)

"

1 3 2x)3 (3x?)3

3 2x 3x2) 3 2x)2(l 3x?) 3 (3x2)

2(1 2x)

1 8x 3 27x 6

3 —2x + 3x2)+ 12x2(1 + 3x2) 27x 4 (1—2x)—3 6as

1—6x + 21x2—4 4x3 + 63x4—5 4x5 - l 27x“.

286 . It is found by observation that the square o f anymultin omial expression may be obtain ed by either o f tworules. Take

,for example, (a b c d)

2. It will be found

that this

and this may be obtained by the following rule ; the squa r eof an y mu ltin om ia l exp r ession consists of the squar e ofeach term

,togeth er with twice th e p r oduct of every p air

of terms.

Again , we may put the result in this form(a b c d)

i

and this may be obtain ed by the following rule ; the squa r eof an y mu l tin om ial exp r ession con sists of th e squa r e ofeach term ,

togeth er with twice the p r oduct of each termby th e sum of all th e terms wh ich f o l low it.

EXAMPLES. XXXI.

1 . (at-

nuts. 2.

3 .

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200 E VOLUTION .

XXXII. Evolution .

287. Evolution is the inverse o f Involution ; so thatEvolution is the method o f finding any propo sed root o fa given number o r expression. It is usual to employ t h eword ex tr act and its derivatives in conn exion with theword r oot; thus, for example, to ex tr act th e squa r e r ootmeans the same thing as to fin d th e squa r e r oo t.In the presen t Chapter we shall begin by stating th ree

simple con sequen ces o f the R ule of S ign s, we shal l thencon sider in succession the extraction o f the roots o f simpleexpression s

,the extraction o f the square root o f compound

expression s and numbers, and the extraction o f the cuberoot of compoun d expression s and numbers .

288. An y even r oot of a p ositive quan tity may beeith er p ositive o r n ega tive.

Thus, for example, a x a : a2, and a x —a = a

g; there

fore the square root o f a2 13 e1ther a o r _ a,that is

,either

289. An y odd r oot of a quan tity has the same signas th e quan tity .

Thus, fo r example, th e cube root o f a3 is a, and the cuberoot of —a

3 is —a .

290. Th er e can be n o even r oot of a n egative qua n tity .

Thus, fo r example, there can be n o square root o f a2;

for if any quan tity be multiplied by itself the resul t isa p ositive quan tity.

Th e fact that there can be n o even root o f a n egativequan tity is sometimes expressed by calling

'

such a root animp ossible quan tity o r an imagin a r y quan tity .

291 . Rule for obtain ing any root o f a simple in tegralexpression . D ivide th e in dex of ever y f ac tor in th e

exp r ession by the n umber den oting th e r oo t, an d give

th e p r op er s ign to th e r esult.

E VOLUTION 201

Thus, for example, J(16a2b4) J(4 2a2b4) 4 4ab”

8a6b90u ) 2/ 23a6b9c‘2) 2a2b3c4 .

sesame) Jew s e a n

292. Rul e for obtain ing an y root o f a fraction . Fin dth e r oo t of the n umer ator and den ominator , and give thep r op er sign to the r esult.

Fo r example,

293 . Suppo se we require the cube root of a2. In thiscase the index 2 is n o t divisible by the number 3 which

the required root ; an d we have, at presen t, n o

other mode o f expressing the result than i/az. Sim ilarly,.Ja , .Ja

s, i/a

‘s, cann ot, at presen t, be otherwise expressed.

Such quan tities ar e call ed sur ds o r ir r a tion a l quantitiesan d we shall cons ider them in the n ext two Chapters.

294 . We n ow proceed to the method of extracting thesqua re root o f a compound expression .

The squa re root o f a2 2ab b2 is a b“

; and we shal l beled to a gen eral rule for the extraction o f the square rooto f any compoun d expression by observing the mann er inwhich a + bmay be derived from a

2+ 2ab b2.

Arrange the terms accord a2 2ab b2 \a 6.

ing to the dimen sion s of on e a2

letter a ; then the first term is2

a2, and its square root is a ,

2“ b) 2ab b

which is the first term o f therequired root. Subtract itssquare, that is a

2, from the whole expr essmn , and bring

down the remain der 2ab+ b2. D ivide 2ab by 2a, and thequotien t is b, which is the other term o f the required root .Take twice the first term and add the secon d term ,

that is,

take 2a + b; multiply this by the second term,that is by b,

and subtract the product, that is 2ab b”, from the remainder. This finishes the operation in the presen t case.

202 E VOLUTION .

If there were more terms we should proceed with a +bas we did formerly with a ; its square, that is, a2+ 2ab+ b2,h as already been subtracted from the preposed expression

,

so we should divide the remainder by 2 (a + b) for a newterm in the root. Then for a n ew subtrahend we multiplythe sum o f 2 (a + b) and th e n ew term

,by the n ew term .

The process mus t be con tinued un til the required rootis found.

295 . Examples.

4x2 12xy 9g2( 2x 3y

4x + 3y) 12xy + 9y2

12xy + 9y2

4x4—2oxs -1- 37x2—30x -l- 9 ( 2x2 5x + 3

4x2 5 x ) 20x ? 3 7x2 3ox 9

2Oxs 25x2

4x2 l ox + 3 ) 12x2—30x + 9

12x2—3 0x - 1 9

x4 4x3y 10x2y2 12xy

3 9314

(x? 2xy

x4

2x2 2xy ) 4x3y 12xy3 9g

4

4x3y'3 4x9y

2

2x2 4xy 3g?) 6x2 2 12xy

3 9g4

12xy3 9y

“

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204 E VOLUTION.

exactly . We sometimes find such an example as the following proposed ; find four terms o f the squar e root of 1—2x .

1—2xkl—x

2—x) —2x

—2x +x2

2 - 2xl

2—2x—x2

x5 s

x3 +x4+—2 4

5 x4 x5

x6

4 2 4

A x5

x6

Thus we have a remainder4 2 4

finding four terms o f the squar e root o f 1—2x ; and so we

x2 5 x4 x

5x6

know that —x —2—2

= 1 —2x +—1

299 . The preceding investigation of the square root o fan Algebraical expression will enable us to demonstratethe rule which is given in Arithmetic for the extraction o f

the square root o f a number.

Th e square root o f 100 is 10, the square root o f 10000is 100, the square root o f 1000000 is 1000, and so on hen ceit follows that

,the square root o f a number less than

must consist of on ly o ne figure, the square root of a

E VOLUTION 205

number between 100 and 10000 o f two places of figur es, o fa number between 10000 and 1000000 o f three places o ffigur es, and so o n . If then a poin t be placed over everysecon d figure in any numb er, beginn ing with the figure inthe un its’ place, the number o f poin ts will shew the n umbero f figures in the square root. Thus , for example, thesquare root. of 4 3 5 6 con sists of two figures, and the squarer oot o f 61 1 5 24 consists o f three figures.

300. Suppose the square r oot of 3249 required.

Poin t th e numb er according to therule ; thus it appears that the rootmust con sist o f two places o f figures.

Let a + b den ote th e root,where a is

the value of the figure in the ten s’ 74 9place, and b o f that in the units’ place.Then a must be the greatest multipleof ten , which has its square less than 3200 ; this is foundto be 5 0. Subtract a 2, that is, the square o f 5 0, from thegiven number

,an d th e remainder is 74 9 . D ivide this re;

mainder by 2a , that is, by 100,and the quotien t is 7,

which is the value o f b. Then (2a + b) b, that is, 107 x 7 o r

74 9, is the number to be subtracted and as there is n own o remainder

,we conclude that 5 0 7 o r 5 7 is the required

square root.

It is stated ab ove that a is the greatest multiple o f tenwhich has its square less than 3 200. Fo r a eviden tly cann o t be a gr ea ter multiple o f ten . If possible, suppo se _

it

to be some mul tiple of ten less than this, say x ; then sin cex is in the ten s’ place, an d b in the un its’ place, x + b is lessthan a ; therefore the square o f x + b is less than a

2,and

con sequen tly x + b is less than the true square root.

If the root con sist o f three places of figures, let a r e

present the hundreds,and b the ten s ; then having ob;

t ain ed a an d b as before,let the hundreds and tens

together be considered as a n ew value o f a, and find a"

newvalue o f b for the un its:

E VOLUTION.

301 . Th e cyphers may be omitted for‘

the sake o fbrevity, and the following rul e may be obtained from theprocess.

Poin t every second figur e, begin n ingwith th at in th e un its’ p lace, an d th us

divide th e wh ole n umber in to p er iods.

Fin d th e gr ea test n umber wh ose squar e 107) 749is con tain ed in th e fir st p er iod ; th is 74 9is th efir stfigur e in th e r oot ; subtr act its

squar e f r om th e fir st p er iod, an d to the

r ema in der br ing down th e n ex t p er iod . D ivide th is

quan tity , omitting th e lastfigur e, by twice the pa r t of the

r o ot alr eady f oun d, a nd a n n ex th e r esu lt to th e r oot anda lso to th e divisor ; th en mul tip ly th e divisor as it n owstands by th e pa r t of the r oot last obtain edf o r th e subtr ah en d. If th er e be m or e p er iods to be br ough t down , th eop er a tion must be r ep ea ted.

302. Examples.

Extract the square root of 132496, and o f 5 322249.

5 322249 ( 23074

In the first example, after the first figure o f the root isfoun d and we have brought down the remainder, we have4 24 ; according to the rule we divide 4 2 by 6 to give then ext figure in th e root : thus apparently 7 is the n extfigure. But o n multiplying 67 by 7 we obtain the product4 69, which is greater than 4 24 . This sh ows that 7 is to olarge fo r the second figure of the root, and we accordinglytry 6 , which succeeds. We ar e liable occasionally in thismann er to try too large a figure, especially at the earlystages of the extraction o f a square root.

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208 E VOLUTION.

3 05 . The fo llowing is _the extraction o f the square roo to f °

4 to Seven dec imal places :

0

3 6

.5 05 76

632425

l 264905 ) 69975 006 324 5 25

12649105 ) 672975 006 324 5 5 25

405 1975

3 06 . We n ow proceed to the method o f extracting th ecube root o f a compoun d expression .

The cube root of a3+ 3a2b+ 3ab3 + b3 is a + b ; and we

sh all be led to a gen eral rule for the extraction o f the cuberoot of an y Compound expression by observing the man n erin which a bmay be derived from a

3+ 3 a2b 3ab2+ b3 .

Arrange th e ‘

terms ao a3+ 3a9b+ 3ab2+ b3 ta + b

cording to the dimen sion so f o n e letter a ; then thefirst term is a 3

,and its cube 3a?) 3a

2b 3ab2+ b3

root is a , which is the first+ 3ab9+ b3term of t h e required root.

Subtract its cube, that isa’

, from the whole expression, and bring down the re;

E VOLUTION . 209

main der 3a’h 3ab2+ b3 . D ivide 3a9b by 3a2, and th e quotien t is b, which is the other term o f the required rootthen subtract 3a2b+ 3abz,

+ b3 from the remainder,and the

whole cube o f a + b has been subtracted. This fin ishes theoperation in the presen t case.

If there were more termswe should proceed with a bas

we did formerly with a ; its cube, that is a3 3a2b 3c 2 b3

,

has already been subtracted from the proposed expression ,

so we should divide the remainder by 3 (a+b)z fo r a n ewterm in the root ; and so on .

3 07. It will be conven ien t in extracting th e cube rooto f more complex expression s, and o f numbers

,to arrange

the process o f the preceding Article in three column s,as follows :

3c +c 3a2

(3a + b) b

3a2 3ab b? 3a2b 3ab2 b3

3a2b 3ab2+b3

Find the first term of th e root, that is a ; put a3under

the given expression in the third column and subtract it.Put 3a in the first column

,and 3a2 in the secon d column ;

divide 3uzb by 3a2, and thus obtain the quotien t b. Add

b to the expression in the first column ; mul tiply the expression n ow in th e first column by b, and place the product in the secon d column , and add it to the expressionalready there ; thus we obtain 3 a2+ 3ab+ b2. Multiplythis by b, and we obtain 3a2b+ 3ab2+ b3 , which is to beplaced in the third column and subtracted. We have thuscompleted the process o f subtracting (a + b)3 from theorigi nal expression . If there were more terms the Operation would have to be con tinued

210 E VOLUTION .

3 08 . In con tinuing the Operation we must add such a;term to the first column , as to obtain there th r ee times thep ar t of th e r oot alr eady f oun d. This is con ven i en tlyeffected thus ; we have already in the firstcolumn 3 a + b ; place 2b below b and add ;thus we obtain 3 a + 3b, which is three timesa + b, that is , three times the part of the rootalready found. Moreover

,we must add such a. 3a + 3b

term to the second column,as to obtain there

th r ee times the squa r e of th e p a r t of th e r oot alr eadyjh u nd . This is conven ien tly effected thus ; we have al readyin the second column and belowth a

ét

1ti a2+ 3ab+ b2 ; place

h

b2

gelowiua

nd (3a + b)bad t e expression s in t e t ree es ; 2

thus we obtain 3a2+ 6ab+ 3b2, which is3“ + 3ab+ bz

three times that is th ree times

tthesquare of the part of the root already 3a? 4. 6ab 4.

enn d.

3 09. Example. Extract the cube root of8x6—3 6x5 102x4 —171z3 + 204x2—1441z+ 64.

—3 :v(6x2—3x)

6x2—9x + 4 1 2w4—18x3 -l- 9x’

l 2x4—3 6x3 + 27x2

—9x + 4 )

12x4—3 6x3 5 1x 9—3 6x + 16

8x6— 3 6x5 + 102x4 171x 3 + 204x2 64 k2x9 4

102x 4 1 7156 3 l 4 4x + 64

5 4x4

4 83 1“ 14 4 :L'3 6 4

4 8x‘64

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EVOLUTION.

a number less than 1000 must consist o f on ly on e figure,the cube root o f a number between 1000 and 1000000 o ftwo places o f figures, and so o n . If then a poin t be placedover every third figure in any number

, beginning with thefigure in the un its’ place, the number o f poin ts will shewthe number o f figures in the cube root. Thus, fo r example,the cube root o f 4 05224 con sists o f two figures, and thecube root o f 12812904 con sists o f three figures.Suppose the cube root o f 274025 requir ed.

180+ 5 10800 274625 ( 60 5

925 216000

1 1725

Point the number according to the rule ; thus it appearsthat the root must con sist o f two places o f figures. Leta + b den ote the root, where a is the value o f the figure inthe ten s’ place, and b o f that in the un its’ place. Then a

must be the greatest mul tiple of ten which has its cubeless than 274000 ; this is found to be 60. Place the cubeo f 60, that is 216000, in the third column under the givennumber and subtract. Place three times 60, that is 180,in the first column

,and three times the square o f 60, that

is 10800, in the second column . D ivide the remain der inthe thir d column by the number in the second column

,

that is,divide 5 8625 by 10800 ; we thus obtain 5

,which

is the value o f b. Add 5 to the first column , and mul tiplythe sum thus formed by 5 , that is, multiply 185 by 5 ; wethus obtain 925 , which we place in the second column and

add to the number already there. Thus we obtain 1 1725multiply this by 5 , place the product in the third column ,and subtract. The remainder is zero, and therefore 6 5 isthe required cube root .Th e cyphers may be omitted for brevity, and the pro

cess will stand thus185

1 1725

E VOLUTION.213

3 11 . Examp le. Extract the cube root of 109215 3 5 ?

1002153 5204 7864

674044

After obtain ing the first two figures o f the root, namely4 7, we adjust the first and second column s in the mann erexplain ed in Ar t. 3 08 . We place twice 7 un der th e firstcolumn

,and add the two lin es, giving 1 4 1 an d we place

the square o f 7 under the second column , and add the lastthree l ines, giving 6 627 . Then the operation is con tinuedas before. Th e cube root is 4 78.

In the cour se o f working this example we might haveimagin ed that the second figur e o f the root would be 8 oreven 9 ; but o n trial it wil l be found that these numbersar e too large . As in the case o f the square root, we are

liable occasionally to try to o large a figure, especially at theearly stages of the Operation .

3 12. Example. Extract the cub e root of 865 3002877.

865 3002877k205 3

s

123025

12625 95 9

In this example the student should n otice the occur~

ren ce of the cypher in the r oot.

214: E VOLUTION.

3 13 . If the root have any number o f decimal places,the cube will have thrice as many ; and therefore the number o f decimal places in a decimal number

,which is a

perfect cube, an d in its simplest state, will n ecessarily be amultiple o f th r ee, and the number o f decimal places in thecube root will n ecessarily be a third of that number. Henceif the given cube number be a decimal

,we place a point

over th e figur e in th e un its’

p lace, and over every thirdfigure to the right and to the left o f it, and proceed as in

the extraction o f the cube root o f an in teger ; then thenumber o f pomts in the decimal part o f the proposednumber will indicate the number o f decimal places in thecube root.

Example. Extract the cube root o f 14 102 327296 .

141023 27296l24'16

s

173 5 21

17467716

3 15 . If any number, in tegral o r decimal , h as n o exactcube ro o t, we may an n ex cyphers, an d proceed With theapproximation to the cube root to any desired exten t.

Th e following is the extraction o f the cube ro ot of ‘

4 to

tour decimal places

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216 EXAMPLES. XXXII.

17. 18. 1

19. 20. w4

21 . 1—4x + 10x2—12x3 + 9x4.

22. 4x8—4a s 7x 4 + 4x2+ 4 .

23 . x4 2ax3 5 d

2x24a 3x 4a4 .

24 . as“ 2ax3 (a

2 2b‘3)ee2 2ab2x b".

25 . x6 l 2a

b+240x

264 .

26. as 4aw5 l oa3m3 4a5x a

“.

27. 1—2x + 3x2 - 4x3 + 5 x4—4x5 + 3x6—2x7+w8

to 9g2

9y2

2: 1 5y5+1 6z

2 +

Find th e fourth roots o f the following expression s

1 4x 6x2 4x3 $

4.

3 0. 16a:4 96x3y 216x2y2 216xy

3 81g“.

3 1 . 1 4x + 10x2 19x4 10w6—4x7+a; 8

3 2. {x4 2(a b)a:3 (a2 4ab b2)x2 2ab(a b)a: a

2b2}2

Find the eighth roots o f the following expression s

3 3 . 6278+ 8337 + 5 6x

5+ 70x4 + 5 6x3 + 28$2+ 827+ 1 .

3 4 . {x4 3x2y

z 2a'y3

Find the square roots of the following numbers3 5 . 1 15 6. 3 6. 2025 . 3 7. 3 721 . 3 8. 5 184 .

3 9. 75 69 . 4 0. 9801 . 4 1 . 1 5 129. 4 2. 10304 1.

4 3 . 165 64 9. 4 4 . 4 5 . 4 12 164 .

4 6 .

'83 5 3 96. 4 7. 1 5 2275 6 . 4 8 . 293 76400.

EXAMPLES. XXXII. 217

4 9. 3 84 5 24 01 . 5 0. 4 981 5 364 . 5 1.

5 2.

‘

24 373 969. 5 3 . 14 4 16804 9. 5 4 . 25 40764 83 6 .

5 5 . 3 2 5 5 13764 . 5 6. 4 5 44 99761 .

5 7.“

5 6875 7305 6. 5 8. 1965 4060224 1 .

Extract the square root o f each o f the following numbers to five places o f decimals :

5 9.

‘

9. 60. 62 1 . 6 1 .‘

4 3 . 62. 0 085 2.

63 . 17. 64 . 129. 6 5 . 3 472 5 9. 66.

Find the cube roots o f the following expression s :67. 8x3 3 6x2y 5 42 g

2 27y3.

68 . 1728x4373 5 76x2y

6

69 . {L3 3x2(a b) 3 .i; (a b)

2(a b)

3.

70. 1 .

71 . w6—3M 5 5 a3x3—3a 5x — as.

72. 4 86x5

806 3013 9064702 108022: 276°

73 . 1 —144x 5 + 64x 6.

74 . 1—3a:+ —102 3 12x 5 —6x7+ 3$8—x9

Find the sixth roots of the following expressions75 . 1 + 12x + 60x2+ l 60x3 + 24Ox

4+

76. 729x6 14 5 8x5 + 1215x4 1827+ 1 .

Find the cube roots of the following numbers77. 19683 . 78. 4 2875 . 79. 1 5 74 64 .

80. 226981 . 81 . 6814 72. 82. 778688 .

83 . 2628072. 84 . 324 1792. 85 . 5 4 0101 5 2.

86 . 6023 62 88. 87. 88. 2 203 48864 .

89. 13 713 3063 1 . 90. 209105 18875 .

91 . 91398648463 125 . 92. 5 340104393239.

218 IND ICES

xxxn r.“

In dices.

3 16 . We have defin ed an index o r exp on en t in Art . 16,and, according to that defin ition , an index has hithertoalways been a positive whole number. We are n ow aboutto extend the defin ition o f an index, by explain ing themean ing o f fractional indices and of n egative indices.

3 17. If m‘

and n'

are an y p ositive who le n umber samx a

nz a

m+n.

Th e truth o f this statemen t, has already been shewn

in A

fit . 5 9, but it is conven ien t to repeat the demonstra

1on ere.a x a x a x to m factors

,by Art . 1 6 ,

a x a x a x te n facto rs, by Ar t. 1 6 ;

thereforex a x a x a x to m+ n f actors

am“

,by Art. 1 6.

In like mann er, ifp is also a positive whole number,a"x a

”x xap z a

mfl l p;

and so on .

3 18. If m and n are positive whole numbers, and mgreater than n ,

we have by Art . 3 17am—n

X a”: a

m -

n -l 'n

therefore

This also has been already shewn see Art. 72 .

3 19. As fractional indices and n egative indices haven ot yet been defined wwe ar e at liberty to give what

.

defini

tions we please to them ; and it is found convement to

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220 IND ICES.

m

3 21 . Requir ed th e mean ing Qf a" wher e m and n ar e

any p ositive wh ole n umber s.

By supposition ,m "

+3+Z

a”x a

”x a

"x to n factor s= a "

therefore a"i must be equivalent to the n th root of a“

,

that is, a

11;

Hen ce a" mean s the nth root o f the m ill power of a ;

that is, in a fractional in dex the numerator den otes a powerand the den ominator a root.

322. We have thus assign ed a mean ing to any positiveindex

,whether whole o r fractional ; it remains to assign a

mean ing to n egative indices.Fo r example, required the mean ing o f a

‘ fl.

By supposition , a3x a

‘ ”413 - 2

41

a,

therefore a 1

We will n ow give the defin ition in gen eral symbols.

323 . R equir ed th e mean ing of wh er e n is any

p ositive n umber wh ole o r f r action al.By supposition, whatever m may be , we ar e to have

amx

Nowwe may suppose m positive and gr eater than n,

and then,by what h as gon e before, we have

a'"

x a": and therefore

Therefore a'"x a

‘ "

therefore

IND ICES . 221

In order to express this in wo rds we will defin e th eword r ecip r oca l. On e quan tity is said to be the r ecip r oca l o f an other when the product o f the two is equal to

1

.v

Hence a ‘ " is the reciprocal of a”

; or we may put thisresult symbolically in any o f the following ways,

1 1a

un ity ; thus, for example, a: is the r ecip r ocal of

x a‘ "

3 24 . It will follow from the mean ing which h as beengiven to a n egative in dex that am when m is lessthan n

,as well as when m is greater than n . For suppose

m less than n ; we have

a

Suppose m = n then a'"

a” is obviously z l ; and

Th e last symbol has n o t hitherto received a

mean ing, so that we ar e at liberty to give it the meaningwhich n atural ly presen ts itself ; hence we may say that

3 25 . In order to form a complete theory o f Indices itwould be n ecessary to give demon stration s o f several proposition s which will be found in the larger Algebra. Butthese proposition s follow so n aturally from the definition sand the properties o f fraction s, that the studen t will n o tfind any difficul ty in the simple cases which will come before h im. We shall therefore refer for the complete theoryto the largerAlgebra, and on ly give here some examples asspecimen s.

3 26 . If m and n are positive whole numbers we kn owthat see A r t . 279. Now this result will alsohold when m and n ar e n o t positive whole numbers. Forexample,

‘For let then by raising both sides to thefourth power we have a5 = .v

4; then by raising both sides

IND ICES.

to th e third power we have a x” therefore .v =a

’fl’i, which

was to be shewn .

327. If n is a positive whole number we kn ow thatThis resul t will also hold when n is n o t

a positive whole number. For example, aix

For if we raise each side to the third power,we obtain in

each case ab ; so that each side is the cube root of ab.

In like mann er we have1 l 1a"

Suppose n ow that there ar e m o f these quan titiesa , b, and that all the rest are equal to a ; thus weobtain

1 1

that is, Z/a)“ 2/

Thus the on “ power of the n th root o f ais equal to thenth root of the m l" power of a .

328 . Since a fraction may take differen t forms withoutany change in its value, we may expect to be able to givedifferen t forms to a quan tity with a fractional index, without altering the value of the quan tity. Thus

,for example,

‘

2 4

3 6

case . Fo r if we raise each side to the sixth power,we

o btain a“; that is, each side is the s ixth root o f a“.

we may expect that a%= al

and this is the

3 29. We will n ow give some examples of Algebraicaloperations involving fractional and n egative expon ents.

Mul tiply all bgcl by a

3b3 c§

therefore all bi cl x a

‘l bl c% a i7l 6 17 6 .

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EXAMPLES . XXXIII .

EXAMPLES. XXXIII

Find the value o f

1 . 9i

. 2. 4 3 3 . 4 . 5 . (an- i .

7. 8 . .Ja“ 9. i/a

‘

10. all x ail x a " l

Mul tiply1 1 . p

lu g/i by x

i—yl. 12.

1 3 . x + xé+ 2 by x + x3 - 2.

14 . by w‘ 4

1 5 .

-l + 1 by a- li— l .

16 . ai—2 -t

'

a

-

3 by a%—a‘ %

17. a + a3b5—a’

y%by a + a3bé+xly§

18 . tel —

.vyl? h ar

dy—yg by x +w3y

3+ y .

D ivide19 . fi —

y’fi by x

i—yi

. 20. a—b by al bl

21 .27y

‘ 2 by 4x" )5 3y

'%

22. .i'g—rlfgl

‘l e d dy—y§ by ail—y

23 . a%+ alibé b%by a?

82 alibi bl .

24 . a% b% c

"% 2a3bl by dl bi 0

132

25 . xi M ing a3 by ad 2aéxé a .

26. Ga'iygt «l ady

gyéby xi Qxiy

iy}

Find the square roots of the following expression s :

27. .v~ 28 . (x + .6 —4 (se—:v

- l).

29. xi? 2l mg.

3 0. 123 3 25 16x' 8

S URD S. 225

XXXIV. Sur ds .

3 30. When a root of a number cannot be exactlyobtain ed it is call ed an ir r a tion al quan tity , o r a sur d .

Thus, for example, the following ar e surds2 3 3

N/‘

é: 11 :

And if a root o f an algebraical expression cann ot beden oted without the use o f a fractional index

,it is also

called an ir r a tion a l quan tity o r a sur d. Thus, for ex

ample, the following ar e surds

.Ja, 32, s/a

s,

Th e rules for operation s with surds fol low from thepropositions o f the preceding Chapter ; and the presen tChapter con sists almost en tirely of the application of thoseproposition s to ari thmetical examples.

3 3 1 . Numbers o r expression s may occur in the f ormof surds , which ar e n o t r ea l ly sur ds. Thus, for example,J9 is in the form of a surd

,but it is n o t really a surd

,fo r

J9 z 3 ; and is in the form o f a surd,but

it is no t really a surd, for J (a2 2ab a b.

3 32. It is often conven ien t to put a ration al quan tityin to the form o f an assign ed surd ; to do this we raise thequan tity to the power corresponding to the root indicatedby the surd, and prefix the radical sign . Fo r example

,

3 =J32: J9 ; 4 :

N3

/4 3= £764 ; a : $614 ; a + bz f/(a - l- bf

3 3 3 . Th e product o f a ration al quan tity and a surdmay be expressed as an en tire surd, by reducing theration al quan tity to the form of the surd, and then multiplying ; se e A rt . 327. Fo r example, 3 J2 z J 9 x

2/8 x J4 z a z Jazx

3 3 4 . Conversely, an en tire surd may be expressed as

the product of a rati on al quan tity and a surd, if the root o fo ne factor can be extracted.

S URD S .

For example, .J32= J(16 x xJ2= 4 N/z;

f/48= 2/8 x 2/6 z 2 i/6 ;

{M3 x gb

2z a '3/b

2.

3 3 5 . A surd fraction can be transformed in toequivalen t expression with the surd part in tegral.

For example,

1

6

6

£72 £718

3 x 9

3 3 6 . Surds which have n o t the same index can be

tran sformed in to equivalen t surds which have ; see Ar t . 3 27

Fer example, take .J5 and i/l l J5 = 5%7 f/l l z fll )

9 5 3 :

3 37. We may n otice an application of the pr ecedingArticle . Suppose we Wish to kn ow which is the greaterJ 5 0 1 .f/1 1 When we have 1 educed them to the sameindex we see that the f01mer is the greater, because 125 isg1 eate1 than 121 .

3 38. Surds ar e said to be simila r when they have, orcan be reduced to have, the same irration al factors.

Thus 4 J7 and 5 J7 ar e similar surds ; 5 i/2 and 42/16ar e also similar surds, for 4 6 z 8 i/2.

3 3 9. To add o r subtract similar surds, add o r subtracttheir coefficien ts

, and afiix to the resul t the commonirrational factor.For example,

4

4 3 3

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SURD S.

3 44 . The only cas e o f division by a compound surdwhich is o f any importan ce is that in which the divisor isthe sum o r differen ce of two quadr a tic surds , that is, surdsinvolving square roots. Th e division is practically effectedby an importan t process wh ich is called r a tion al is ing th e

den omin a tor of a f r action . For example, take the fraction4 0

5If we multlply both nume1 at01 and den omi

nator o f this fraction by 5 .J2—2J3 , the value o f the frac;tion is n o t altered, while its den om in a tor ismade r a tion a l ;

4

5 J2+ 2 J3“

(5 J2 + 2 J3 ) (5 J2—2 J3 )10 9 2—4 J3

5 0—12 1 9

2 3 (2 N/3 (2 .J3 + J?)8 + 3 J6 8 + 3 J612—2 1 0

3 4 5 . We shall n ow shew how to find the square root o fa binomial expression , on e o f whose terms is a quadraticsurd. Suppose

,for example

,that we require the square

root o f 7 + 4 J3 . Sin ce (J r -r 2 it isobvious that if we find values o f a: an d y from x + y = 7,and 2 J(xy)= 4 J3 , then the square root o f 7 + 4 J3 will be4 37+ N/y . We may arrange the whole process thus

Suppose .J(7+ 4 J3 )= .Jx Jy ;

square,

Assume .v y ==7, then 4 J3 ;

square,and subtract

, (a: y)2 4xy 4 9 48 1 ,

Similarly,

that is, (.v—y )

2= 1,therefore .v—y = 1 .

Sin ce .v+ y = 7 and .v—y z l,we have .v= 4

, y = 3

therefore J3 = 2 +V3

Similarly, (7 4 J il l 2 Q3 .

EXAMPLES . XXXI V 229

EXAMPLES. XXXIV.

Simplify1 . 3 .J2 + 4 ,J8 4 32. 2. mos.

3 . 2 .J3 3 4 .

Mul tiply

5 . J 5 + nag) 55 by

6 .—J

1

16- l-

3

1

2by

7. 1 + J2 by J6 — J2.

8 . .J3 + J2 by

Rationalise the den omin ators o f the foll owing fractions

2—4 2‘

J2'

Js 2J3 + 3 J2

Extract the square root o f

13 . 14 + 6J 5 . 14 . 1 6—6 .J7.

1 6 . 4 —J1 5 .

Simplify1 1

“OH M/3 )20. N/5 )

1 + . l3

XXXV. Ratio.

3 4 6. Ratio is the relation which o ne quan tity bearsto an other with respect to magnitude, the comparisonbeing made by considering what multiple, par t, o r parts,the first is o f the second.

Thus, for example, in comparing 6 with 3 , we observethat 6 h as a certain magn itude with respect to 3 , whichit contain s twice ; again , in comparing 6 with 2, we see that6 h as n ow a different r ela tive magn itude, for it con tains2 three times ; o r 6 is greater when compared with 2 thanit is when compared with 3 .

3 47. Th e ratio o f a to b is usually expressed by twopoin ts placed between them,

thus , a b ; and the former iscalled the an teceden t o f the ratio, and the latter the cousequen t of the ratio.

3 48. A ratio is measured by the fraction which h as forits numerator the an teceden t o f the ratio, and for itsden ominator the con sequen t o f the ratio . Thus the ratioo f a to b is measured by g then for shortn ess we maysay that the ratio o f a to b is equal to E

b

bo r is

3 49. Hence we may say that the ratio of a to bis equal

to the ratio of c to d, whena 0

3 5 0. [ f th e terms of a r atio be m ultip lied o r dividedby th e same qua n tity the r a tio is n o t al ter ed .

(Ar t.

3 5 1 .

.

We comp ar e two or more ratios by reducingthe fractl o ns which measure these rat ios to a commonden ominator. Thus, suppose o ne ratio to be that o f a to b,

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232 RA TIO .

the ratio ac bd is said to be compounded o f the two ratiosa b and c d.

When the ratio a b is compounded with itsblf theresulting ratio is a2 bZ ; this ratio is sometimes called thedup licate ratio o f a b. And the ratio a3 b3 is sbmetimescalled th e tr ip lica te ratio of a b.

3 5 6. Th e following is a very importan t theorem con

cern ing equal ratios.

Suppose that then each o f these ratios

where p , q, r'

, n are any numbers whatever.

a c eFo r let h

f;

hbz a,kd z c, hf z e ;

therefore p (hb)"

qHad)" r (kf )" =p a

"

qc"

r e”

_p a

"

qc"+ r e

"

therefore —p b

"

qd"

rfn )

therefore

Th e same mode ofdemon stration may be applied, anda similar result obtain ed when there ar e more than th r ee

ratios given equal .As a particular example we may suppose n = 1, then we

see that if Z each of these ratios is equal to

w and then as a special case we may supposep b qd r)

“

p : q z r,so that each o f the g1ven equal ratios 13 equal to

a + c + e

b+d+f°

EXAZlIPLES . XXXV 233

EXAMPLES. XXXV.

1 . Fin d the ratio of fourteen shillings to three guin eas.

2. Arrange the following ratios in the order o f magn itude ;

3 . Find the ratio compounded o f 4 1 5 and 25 3 6 .

4 . Two numbers are in the ratio o f 2 to 3 , and if 7 beadded to each the ratio is that of 3 to 4 : find the numbers.

5 . Two numbers ar e in the ratio o f 4 .to 5 , . and if 6 be

taken f1 om each the ratio is that o f 3 to 4 : find the numbers.

6 . Two numbers ar e in the ratio of 5 to 8 ; if 8 beadded to the less numbe1

,and 5 taken f1 om the greate1

numbei,the ratio is that o f 28 to 27 : find the numbers.

7 . Fin d the number which added to each term o f theratio 5 3 makes it th ie e- fo rut h s of what it would have become if the same number h ad been taken from each term .

8 . Find two numbers in the ratio o f 2 to 3 , such thattheir difi'eren ce h as to the differen ce o f their squares theratio o f 1 to 25 .

9 . Find two numbers in the ratio of 3 to 4, such that

their sum has to the sum o f their squares the ratio of7 150 5 0.

10. Find two numbers in the ratio of 5 to 6, such thattheir sum has to the differen ce o f their squares the ratio o f1 to 7.

1 1 . Find .v so that the ratio x : 1 may be the duplicateo f the ratio 8 : .v .

1 2 . Find a: so that the ratio a—x b—x may be th edupl icate o f the ratio a : b.

1 3 . A person h as 200 coin s con sisting o f guin eas, hal fsovereigns , and half- crowns ; the sums ofmon ey in guin eas,hal f—sove1 e 1gns , and half- crown s are as 3 ; findthe n umbers o f the differen t coin s.

1 4 . If b —b 6a—b,find a b.

1 5 . IfI

a—b b—c c a’ 0.

PROP018TION .

XXXVI. Pr op or tion .

3 5 7. Four numbers are said to be proportional whenthe first is the same multiple, part, or parts of the second

as the third is of the fourth ; that is when th e fournumbers a, b, c, d ar e called proportionals. This is usuallyexpressed by saying that a is to b as c is to d ; and it isrepresented thus a b c d

,o r thus a b= c d.

The terms a and d are called the ex tr emes, and b and c

the mean s.

3 5 8. Thus when two ratios are equal, the four numberswhich form the ratios are called proportionals ; and the present Chapter is devoted to the subj ect of two equal ratios.

3 5 9 . Wh en f our n umber s a r e p r op or tionals the p r oduct of th e ex tr emes is equal to th e p r oduct of th e mean s.

Let a, b, c, d be proportionalsa c

b d’

multiply by bd thus ad z be.

If any three terms in a proportion are given , the fourthmay be determin ed from the relation ad be.

If b= c we have ad= b2 that is, if th e fir st be to the

second as th e secon d is to th e th ir d,th e p r oduct of th e

ex tr emes is equal to th e squar e of th e mean.

When a : b b d then a, b, d are said to be in co n

tin uedp r op or tion ; and b is called the mean p r op or tiona lbetween a and d.

3 60. If th e p r oduct of two number s be equal to th e

p r oduct of two oth er s, th e f our a r e p r op or tionals , th eterms of eith er p r oduct being taken f o r the means, andth e terms of th e oth er p r oductf or th e ex tr emes.

For let xy z ab; divide by ay , thusx b

or x za° ° b zy

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PROPORTION .

3 66. A lso thefir st is to the excess of thefir st above theseco n d as th e th ir d is to th e excess of th e th ir d above thefour th .

By the last Article“

Eb 0

5d

a—b b c—d dthereforeb

x

a dx

E’a

o r a—b a e— d 0 ; therefore a a —b c c— d.

3 67. Wh en f our n umber s a r e p r op or tion a ls , the sumof the fir st an d secon d is to th eir difi

'

er en ce as th e sumof th e th ir d an d f our th is to th e ir difi

’

er en ce ; that is, ifa b e at

,then a + b a —b c + d c— d.

a + b c+ d a —b c—d- By Arts. 3 64 and 3 65 —

b d b d

a + b a—b c + d c — d a + b c+ dthereforeb b d d

that 13a

_

—b c d’

or a + b : a —d

3 68 . It is obvious from the precedin g Articles that iffour numbers are proportion al s we can derive fr om themmany other proportion s ; see also Art . 3 5 6.

3 69 . In the defin ition ofProportion it is supposed thatwe can determin e what multiple or what part o n e quan tityis o f an other quan tity o f the same ki nd. But we can n otalways do this exactly. For example

,if the side o f a

square is o n e in ch long the length o f the diagonal is den oted by ,J2 in ches ; but ,J2 cann ot be exactly foun d, sothat the ratio o f the length of the diagon al of a squareto the length o f a side cann ot be exactly expressed bynumbers. Two quan tities ar e called in commen sur able

when the ratio o f o n e to the other cann ot be exactly expressed by numbers.

The studen t ’s acquain tan ce with A rithmetic will suggest to h im that if two quan tities ar e real ly in commensurable still we may be able to express the ratio o f o n e tothe other by numbers as n early as we please . For example

,

we can fin d two mixed numbers, o n e less than 2,and the

other greater than V2, an d o n e differing from t 0 other byas small a fraction as we please.

PROPORTION . 237

370. We will g ive on e propositionwith respect to thecomparison of two in commen surable quan tities .

Let x and y den ote two quan tities ; and suppose itkn own that howe ver great an in teger q may be we can find

an other integer p such that both x and y lie between13 pi

p

—14: then x and y ar e equal .

Q 9

For the differen ce between x and 31 cann ot be so greatas I ; an d by taking q large en ough 1 can be made lessq Q

than any assign ed quan tity whatever. But if x and y wereun equal their difference could n o t be made less than anyassigned quan tity whatever. Therefore x and y must beequal

3 71 . It will be useful to compare the defin ition of pr oportion which h as been used in this Chapter with thatwhich is given i n the fifth book o f Euclid. Euclid’

s defin i

tion may be stated thus : four quan tities ar e proportionalswhen if any equimultiples be taken of the first and thethird

,and also any equimultiples of the second and the

four th,the multiple of the third is greater than , equal to,

o r less than ,the mul tiple o f the fourth, according as the

multiple of the first is greater than, equal to, o r less thanthe multiple of the second.

3 72. We will first shew that if four quan tities satisfythe algebraical defin ition o f proportion

,they will al so

satisfy Euclid’

s .

For suppose that a b c d ; then therefore

53 52, whatever numbers p and q may be. Hen ce p e is

greater than , equal to, o r less than qd , according as p a isgreater than , equal to, or less than qb. That is

,the four

quan tities a , b, c, d satisfy Euclid’

s defin ition of proportion .

3 73 . We shall n ext shew that if four quan tities satisfyEuclid’

s defin ition o f proportion they will also satisfy thealgebraical defin ition .

For suppose that a , b, c, d ar e four quan titi es such th atwhatever numbers p and q may be, p c is greater than ,

PROPORTION .

equal to , or less than get, according as p a is greater than ,equal to, o r less than qb.

F irst suppose that c and d ar e commensur able ; takep and q such that p c z qd then by hypothesis p aq : thusp a 739 ther efore a

Therefore a b e d.

qb qd

Next suppose that c and d are in commensur able.

Then we can n ot find whole numbers p and q, such thatp c z qd . But we may take any multiple whatever o f d, asqd , and this will lie between two con secutive mul tiples of c,say between p e and Thus _

p_

cis less than un ity,

qd

is greater than un ity. Hence,by hypothesis,

(1 (p+ l ) a

IS less than un ity, anqb

13 greater than un ity.

and(

itare both gr eater than and both less than

qAnd since this is true however great p and q may

be, we in fer that Zand 5 cannot be un equal ; that is, theymust be equal : see Ar t. 3 70. Therefore a b c d.

That is, the four quan tities a , b, c, d satisfy the algobraical defin ition o f proportion .

3 74 . It is usually stated that the Algebraical definitiono f proportion cannot be used in Geometry because there isn o method o f representing geometrically the resul t of theOperation o f division . Straight lin es can be represen tedgeometrically, but n o t the abstract n umber which expresseshow often on e straight lin e is con tain ed in an other. But itshould be observed that Euclid ’

s defin ition is rigorous andapplicable to in comm en sur able as well as to comme ns u rable quan tities ; while the Algebraical defin ition is, strictlyspeaking, confined to the latter. Hence this considerationalon e would furn ish a sufficient reason for the defin itionadopted by Euclid.

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XXXVII. Var ia tion .

3 75 . Th e presen t Chapter con sists o f a series o f pr opositions conn ected with the defin itions o f ratio and proportion stated in a n ew phraseology which is

'

conven ien tfor some purposes.

3 76 . On e quan tity is said to vary dir ectly as an otherwhen the two quan tities depend on each other

,and in such

a mann er that if o n e be changed the other is changed inthe same proportion .

Sometimes for shortn ess we omit the word dir ectly ,and say simply that o n e quan tity varies as an other.

3 77. Thus, for example, if the altitude o f a triangle beinvariable

,the area varies as the base ; for if the base be

increased o r dimin ished,we kn ow from Euclid that the

area is in creased or dimin ished in the same proportion .

We may express this result with Algebraical symbo ls thus ;let A and a be n umber s which represen t the areas o f twotriangles having a common altitude

,and let B and b be

n umber s which represen t the bases o f these triangles r e

spectively ; then 4B

a b

g_ by Art . 3 63 . If there be a third trian g le having thesame altitude as the two already con sidered, then the ratioo f the number which represents its area to the number whichrepresents its base will also be equal to

(

5

1°

ill

- : m,and A s . Here A may repres en t theB

area of an y on e of a ser ies o f triangles which hav e a co n1~

mo n altitude, and B the corresponding bas e, an d m r e

main s con stan t. Hen cc the statemen t that the ar ea variesas the base may also be expressed thus , the ar ea has a

And from this we deduce

VARIA TION . 241

constan t ratio to the base ; by which we mean that th en umber which represen ts the area bears a constan t ratioto the n umber which represents the base.These remarks ar e intended to explain the n otation an d

phraseology which ar e used in the presen t Chapter. Whenwe say that A varies as B ,

we mean that A represents thenumerical value o f any on e of a certain series o f quan tities,and B the numerical value o f the corresponding quan tityin a certain other series, and that A s

,where m is

some n umber which remains con stan t for every corresponding pa1r of quant l ties.

It will be conven ien t to give a formal demon stration

If the relation A s

,deduced from the defin ition in

rt. 376 .

3 78. If A var y as B,th en A is equal to B multip lied

by some con stan t n umber .

Let a and b den ote on e pair of corresponding values o fthe two quan tities, and let A and B den ote any other pair ;

141 Q by defin ition . Hence A —

l B=mB , Wherea b b

m is equal to the constan t(

5.

3 79. Th e symbol 00 is used to express variation ; thusA oc B stands for A var ies as B .

3 80. One quan tity is said to vary in ver sely as another,when the first varies as the r ecip r ocal o f the second. SeeAr t. 3 23 .

O r 1fA _

B ’where m IS constan t, A 13 sa1d to vary

inversely as B .

3 81 . On e quan tity is said to vary as two othersjoin tly ,when , if the former is changed in any mann er, the producto f the other two is changed in the same proportion .

O r ifA =mB 0,where m is con stan t, A is said to vary

Jointly as B and C.

r. A.

VARIATION.

3 82. One quan tity is said to vary dir ectly as a secondand in ver sely as a th ird, when it varies j ointly as thesecond and the reciprocal of the third.

O r ifA =m

0,

B,w

directly as B and inversely as 0.

here m is con stan t, A is said to vary

383 . If A 00 B, and B a c, th en A a e .

Fo r let A =mB ,and B = n 0

, where m and n ar e con

stan ts then A =m n C ; and, as mn is constan t, A at C.

3 84 . If A oc C, and B 00 0, th en A=E B cc C, and

4 (AB) cc C.

Fo r let A =m0’, and a C, where m and n are con

stan ts ; then therefore cc 0.

Also J(AB ) J (m720 2) therefore N/(AB ) cc (7.

385 . 17 11 a BC, th en B cc and C oc -

B

Fo r let A =mBO’

, then B therefore B cc

Similarly, 0 00

3 86. If A at B,an d C cc D ,

then AC at: BB .

Fo r let A = mB,and C= n D ; then AC=mnBD ;

therefore A O cc B D .

Similarly, if A at B,and (

Joe D

, and E ccF, thenACE cc BD F; and so on .

3 87. E A of: B,th en A“

cc B“

For lot A =mB,then therefore A ”

at: B”.

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EXAMPLES. XXXVI ] .

EXAMPLES. XXXVII.1 . A varies as B , and A = 2 when B = 1 ; find the

value o f A when B = 2 .

2 . If A 2 B 2 varies as A 2 B 2, shew that A +B

varies as A—B .

3 . 3A 5B varies as 5A 3B , and A = 5 when B = 2 ;find the ratio A B .

4 . A varies as n B C ; and A = 4 when B’

r and

C= 2 ; and A = 7 when B = 2, and find n .

5 . A varies as B and 0 j oin tly ; and A = I whenB = 1, and find the value o f A when B = 2 and

A varies as B and C j oin tly ; and A = S whenB = 2

,and find the value of B C when A 10.

7. A varies as B and C j oin tly ; a nd A = I2 whenB = 2, and find the val ue o f A B when

8 . A varies as B an d C j ointly and A a whens , and find the value of A when B : b2 an d

C : 02.

9 . A varies as B directly and as C inversely ; andA a

when B =b, and find the value of A when B - c and

Uz b.

10. Th e expen ses o f a Charitable Institution ar e partlycon stan t, and partly vary as the n umber o f inmates.

When the inmates ar e 960 and 3000 the expen ses are r e

spectively £ 1 12 and £ 180. Find the expenses for 1000inmates.

1 1 . The wages of 5 men for 7 weeks being £ 17. 10s.

find how many men can be hired to work 4 weeks for £ 30.

12. If the cost o f making an embankmen t vary as thelength if the area o f the tran sverse section and height beconstan t

,as the height if the area o f the tran sverse section

and length be con stan t,and as the area o f the tran sverse

section if the length and height be constan t, and an em

bankment 1 mile long, 10 feet high , and 12 feet broad cost£ 9600 find the cost o f an embankmen t half a mile long,16 feet high, and 1 5 feet broad.

ARITHMETICAL PROGRESSIOM 245

XXXVIII. Ar ithmetical Pr ogr ession .

3 91 . Quantities are said to be in Ari thmetical Pr ogr ession when they in creas e o r decrease by a common differen ce.Thus . the following series are in Arithmetical Pro

gression ,

14,

a,a + b, a + 2b, a + 3b, a + 4b

The common differen ce is foun d by subtract ing anyterm from that which immediately follows it. In the firstseries the common differen ce is 3 ; in the second series it is2 ; in the third series it is b.

3 92. Let a denote the first term o f an A rithmeticalProgression , b the common differen ce ; then the secondterm is a + b, the third term is a + 2b, the fourth term isa+ 3b, and so on . Thus the n th term is a + (n — l ) b.

3 93 . To find th e sum of a given n umber of terms ofan A r ithmetica l Pr ogr ession , th e fir st term an d the comm on dif er en ce being supp osed kn own .

Let a den ote the first term,6 the common difference

,it

the number of terms,I the last term, s the sum o f the

terms . Thens= a +

And,by writing the series in the reverse order, we have

alsos : l + (l — 2b)+

Therefore,by addition ,

to n terms

therefore s

246 ARITHMETICAL PROGRESSION.

—1)b

— l )b}

Th e equation (3 ) gives the value o f s in terms of th equan tities which were supposed kn own . Equation (1 ) alsogives a conven ien t expression for s

,and furn ishes the

following ru le : the sum of an y n umber of terms in

A r ithm etica l Pr ogr ession is equa l to th e p r oduct of th e

n umber of th e terms in to h alf th e sum of the fir st an dlas t terms .

We shall n ow apply the equation s in the presen t Articleto solve some examples relatin g to Arithmetical Progr ession .

394 . Find the sum o f 20 terms o f the series 1,2, 3 , 4

Here a = 1,b= l n = 2o ; therefore

3 95 . Find the sum of 20 terms o f the series,

5 ,

Here a = 1 b= 2,n : 20 therefore

,

(2 + 19 x 2) x

3 96 . Find the sum o f 12 terms of the series 20, 18,Here a : 20

, b—2

,n 12 ; therefore

(4 0—2 x

3 97. Find the sum of 8 terms o f the series

Here a n 8 therefo re

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EXAMPLES'. XXXVIII .

9 . Insert 7 Arithmetical means between 8 and —4 .

10. Insert 8 Arithmetical means between —1 and 5 .

11. Th e first term of an Arithmetical Progression is13 , the second term is 1 1 , the sum is 4 0 : find the numbero f terms .

12. Th e first term o f an Arithmetical Progress ion is5,and the fifth term is 1 1 find the sum o f 8 terms.

13 . Th e su mo f four terms in Arithmetical Progressionis 4 4 , and the last term is 17 : find the terms .

14 . The sum o f three numbers in Arithmetical Pr ogression is 21 , and the sum o f their squares is 1 5 5 : find thenumbers.

1 5 . Th e sum o f five numbers in Arithme tical Pr ogression is 1 5 , and the sum o f their squares is 5 5 : find thenumbers.

16. Th e seven th term o f an Arithmetical Progressions 12

, and the twelfth term is 7 ; the sum o f the series is1 71 : find the number o f terms.

17. A traveller has a journey o f 14 0 miles to perform.

He goes 26 miles the first day, 24 the second, 22 thethird, and so on . In how many davs does he perform thej ourn ey 2

18 . A sets out from a place and travels 2—5 miles an

hour. B sets out 3 hours after A,and travels in the

same direction , 3 miles the first hour, 35 miles the second ,

4 miles the third, and so o n . In how many hours will B

overtake A l19 . Th e sum of three numbers in Arithmetical Pr o

gression is 12 ; and the sum o f their squares is 66 : findthe numbers .

20. If the sum of n terms o f an Arithmetical Pr ogression is always equal to n 2, find the first term an d thecommon difference.

GEOMETRIC’AL PROGBESSION . 249

XXXIX. Geometr ical Pr ogr ession .

400. Quan tities ar e said to be in Geometrical Pr ogressio n when eac h is equal to the product o f the precedingand some con stan t factor. The constan t factor is calledthe common r atio o f the series, o r more shortly, the r a tio.Thus th e following series are in Geometrical Pr ogr es

sion .

1,3 , 9, 27, 81 ,

1 1 l 1

a,ar

,ar

2,ar

3,ar

“,

1

The common ratio is found by dividing any term bythat which immediately precedes it. In the first example

1the common ratio is 3 , in the second it is Q’in the third

it is r .

401 . Let a den ote the first term of a Geometrical Pr ogr ession , r the common ratio ; then the second term is ar

,

the third term is a r2,the fourth term is a r

3, and so 011 .

Thus the n th term is ar m" .

402. To fi n d the sum of a given n umber of terms of a

Geom etr ica l Pr ogr ession , th e fir st term and th e commonr a tio being supp osed kn own .

Let a den ote the first term,r the common ratio, n the

number o f terms, s the sum of the terms. Then

s = a + ar + a r2 + ar 3 +

therefore sr a r ar2

ar 3 ar’“ 1

ar"

Therefore,by subtraction ,

sr s ar"

a

therefore

250 GEOME TRICAL PROGBESSION.

If l den ote the last term we have

therefore

Equation (1) gives the value o f s in terms o f thequan tities which we1 e supposed kn own. Equation (3 ) issometimes a conven ien t for .m

We shall n ow apply these equation s to solve some examples relating to Geometrical Progression .

4 03 . Find the sum o f 6 terms o f the series 1 , 3 , 9,Here a = 1 , r _ - 3

,n = 6 ; therefore

3 6 1 729—1

3 1 3 13 64 .

404 . Find the sum o f 6 terms o f th e se ries 1,

Here a = 1 , r = —3 , n = 6 ; therefore—1 729—1

3 1182.

405 . Find the sum o f 8 terms of the series 4 , 2,

Here a = 4,r n = 8 therefore

25 5x2 25 5

6 4 1 3 2

4 06 Find the sum o f 7 terms o f the series, 8, 4,

Here a 8 , r n 7 ; therefore

129x2 4 3

16x

3 8

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25 2 EXAMPLES. XXX IX .

4 10. Recurri ng decimals ar e examples o f what are

called infin ite Geometrical Progression . Thus for example3 24 24 24

den otes10 W W

Here the terms after form a Geometr ical Pr ogr es

sion , o f which the first term is and the cemmon ratio

Hence we may say that the sum o f an infin ite

number of terms o f this series is that is

Therefore the value of the recurring decimal is

The value of the recurring decimal may be found practically thus :Let s=

‘

32424 .

then 10 s 3 2 424

and 1000 s 3242 424

Hence,by subtraction , (1000 10) s z 324 3 z 321

3 21therefore990

And any other example may be treated in a similarmann er.

EXAMPLES. XXXIX.

Sum the following series1 1 ,

3 . 25,

4 . 1, g2, 2, 24 2,

EXAMPLES . XXXIX. 25 3

to 6 terms.

to 7 terms.

to infin ity.

to infin ity.

to infin ity.

10. 6,—2 to infinity.

Find the value o f the following recurring decimals1 1 .

‘

1 5 1 5 1 5 12 .

1 3 . 14 .

1 5 . Insert 3 Geometrical means between 1 an d 25 6.

1 6. Insert 4 Geometrical means between 5 4 and 4 03.

1 7. In sert 4 Geometrical mean s between 3 and - 729.

1 8. Th e sum of three terms in GeometricalProgressionis 63 , and the difference of the first and th ird terms is 4 5 :find the terms.

19. The sum o f the first four terms o f a GeometricalProgression is 4 0, a nd the sum o f the first eight terms is3 280 : find the Progression .

20. The sum o f three terms in Geometr ical Progression is 21, and the sum o f their squares is 189 : find theterms.

25 4 HARM ONIOAL PROGRESSION

XL. Harmon ical Pr ogr ession .

4 1 1 . Three quan tities A ,B , C ar e said to be in Har

monical Progression when A C A - B B 0.

Any number o f quan tities ar e said to be in Harmon icalProgression when every three cons ecutive quantities are inHarmon ical Pr ogr ess1on .

4 12. Th e r ecip r oca ls of qua n tities in Ha rmon icalPr ogr ession ar e in A r ithmetica l Pr ogr ession .

Let A ,B , 0 be in Harmon ical Progression then

A : C " A —B B —C.

Therefore A (B C ) C (A —B ).

1 l 1 lD 1V1de by AB C ,

thus5—1—

9 B A

This demon strates the pr epo sitio n .

4 1 3 . The property established in the preceding Articlewill enable us to solve some question s relating to Harmo nical Progression . For example

,insert five Harmon ical

2means between

5 1

8

5. Here we have to mser t five

A r ithmetica l means between 3and Hen ce,by equa

tion (2) of Art . 3 93 ,

therefore 6b therefore b

Hence the Arithmetical Progression is

27 28 2 1 5

1 6’1 6

’13, 8

; and therefore the Harmomcal Pro2 1 6 1 6 1 6 1 6 1 6 8

« r esp

81011 IS3,25 26

,27 23

’29

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25 6 PERM UTATIONS AND COMBINA TIONS .

XLI. Permutations and Combina tion s.

4 15 . Th e differen t orders in which a set o f things canbe arranged are called their p erm utation s.Thus the permutation s o f the three letters a

,b,0,taken

two at a time, are ab, ba, ac, ca, be, ob.

4 16 . Th e combin a tion s of a set o f things ar e thedifferen t col lection s which can be formed out o f them ,

without regarding the order in which the things ar e placed.

Thus the combin ation s of the three letters a,b,0, taken

two at a time,are ab, a c, be ; ab and ba , though differen t

p er muta tion s, form the same combina tion , so also do ac

and ca,and be and ab

4 17. Th e n umber of p er nm ta tions of 11 th ings takenr a t a time is n (n — l ) (n—2) (n

Let there be n letters a , b, c, d, we shall first findthe number of permutation s o f them taken two at a time.Put a before each o f the other letters ; we thus obtainn — l permutations in which a stands first. Put 6 beforeeach o f the other letters ; we thus obtain n—l permutation s in which b stands first. Similarly there ar e n — lpermutation s in which 0 stands first. And so o n . Thus

,

o n the whole, there ar e n (n 1 ) permutation s of n letterstaken two at a time. We shall n ext find the number ofpermutation s of n letters taken th r ee at a time. It h asjust been shewn that o ut of n letters we can form n (n l )permutation s

,each o f two letters ; hence out of the n — l

letters b,c,d,

we can form (n 1) (n — 2) pe rmutations ,each of two letters : put a before each of these

,and

we have (n 1 ) (n —2) permutations, each o f three letters,in which a stands first. Similarly there ar e (n — l ) (n — 2)permutation s, each of three letters, in which b stands first.Similarly there ar e as many in which c stands first. And

so o n . Thus,on the whole, there are n (n — l ) (n - 2) per

mutations o f n letters taken th r ee at a time.

PERM UTA TIONS AND COMBINA TIONS. 25 7

From considering these cases it might be conjecturedthat the number o f permutation s o f n letters taken r at a

time is n (u—l ) (n and we shal l shewthat this is the case. Fo r suppose it kn own that the number o f permutation s o f n letters taken r—l at a time isn (n l ) (n (r l ) we shall shew that a similarformula will give the number o f permutation s o f n letters

,

taken r at a time. Fo r out o f the n —l letters b, c,we can form (n — l ) (n - 2) {n 1—(r 1} pe rmutations

, each‘

o f r —l letters : put a before each of these, andwe obtain as many permutation s, each o f r letters, inwhich a stands first. S imilarly there ar e as many permutation s, each of r letters

,in which b stands first. Simi

lar ly there ar e as many permutation s, each of r letters,

in which 0 stands first.,And so o n . Thus on the whole

there are n (u—l ) (n —r + l ) permutations of n

letters taken r at a time.

If then the formula holds when the letters are taken r 1

at a time it will h e ld when they are taken r at a time.But it h as been shewn to hold when they ar e taken th r ee

time,therefore it holds when they are taken f our at a

time,and therefore it holds when they ar e taken five at a

time,and so o n : thus it holds universally.

4 18. Hence the number o f permutations of n thingstaken all together is n (n 1) (n 1 .

4 19. Fo r the sake o f brevity n (n l ) (n is oftenden oted by Ln thus Ln den otes the product of the naturalnumbers from 1 to n inclusive. The symbol [ a may beread

, factor ial n .

420. Any combin a tion of r th ings will p r oduce [_ rp ermuta tion s.

For by Art . 4 18 the r things which form the givencombination can be arranged in Lr differen t orders.

421; The n umber of combin ation s of 11 things taken rat a time is

n (u - l ) (n — 2) (n - r + l )

25 8 PERM UTA TIONS AND COMBINA TIONS .

For the number o fp ermutation s o f n things taken r at

a time is n (n l ) (n — 2) (n —r + 1 ) by Ar t. 4 17 ; and eachcombination produces li permutations by Art. 4 20 ; hencethe number o f combinations must be

72 (n l ) (n —r + 1 )

l i

If we mul tiply both n umerator and den ominator o fls:this expression by In - r it takes the form

value o f course being un changed

4 22. To fi nd th e n umber of p ermutations of 11 th ings

taken a ll togeth er wh ich a r e n o t a ll dif er en t.

Let there be n letters ; and suppose p o f them to be a,

q o f them to be b, r o f them to be 0, an d the rest o f themto be the letters cl

,e, each occurring singly : then the

number o f permutations of them taken all together will be

Fo r suppose N to represen t the required number o fpermutation s. If in any o n e o f the permutations the 17letters a .were changed in to 1) n ew and differen t lette rs

,

then,without changing the situation o f an y o f the other

letters,we could from the single permutation produce |p

differen t permutations : and thus if the p letters a werechanged in to p n ew and differen t letters the whole numberof permutation s would

_be N x Lg . Similarly if the g letters

b were also changed in to q n ew and differen t letters thewhole n umber o f permutations we could n ow obtain wouldbe N xLg x Q. A n d if the r letters 0 were also changedinto r n ew and differen t letters the whole number of permutation s would be N x U} x [g x (r . But this numbermust be equal to the number o f permutations of n differentletters taken all together, that is tow.

Thus N x [p x lg x lfi Ln ; therefore N

And similarly any other case may be treated.

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BINOM IAL THEOREZII.

XLII. B in omia l T heor em.

424 .

i

We have already seen that (a: 2xa a2,

and that (x a )3 x3 3x2a 3xa2 a

3 the obj ect o f thepresen t Chapter is to find an expression for wheren is any positive in teger.

4 25 . By actual mul tiplication we obtain

(x +

(x +a) (a +b)

(abc bcd cda dab)x abcd.

Now in these results we see that the following lawshold :

I. Th e number o f terms on the right - han d side is on e

more than the number of binomial factors which ar e multiplied together.II. Th e expon en t o f x inthe first term is the same as

the number o f bin omial factors,and in the other terms

each exponen t is less than that o f the preceding term byimity.

III . Th e coefficien t o f th e first term is un ity ; thecoefficien t o f the second term is the sum o f the secondletters o f the bin omial fac tors ; the coefficien t o f the thirdterm is the sum o f the products o f the second letters o f

the bin omial factors taken two at a time ; the coefficien t o fthe four th term is the

,sum o f the products o f the second

letters o f the bin omial factors taken three at a. time ; andso o n ; the last term is t he product o f all the second lettersof the bin omial factors.

We shall shew that these laws always h old, whate verh e the number o f binomial factors. Suppo se the lawsto h eld when n 1 factors ar e mul tiplied together ; that is,

B INOM IAL THE OREM 261

suppose there are n 1 factors a: a,a b, a: c, a: k,

and thatb)

- 2+ u,

where p = the sum o f the letters a , b, c, lo,

q= the sum o f the products o f these letters takentwo at a time

,

r = th e sum of th e products of these letters takenthree at a time,

u = th e product of all these letters.

Multiply both sides o f this iden tity by another factorx + l, and

parrange the product on the right hand according

to powers o f x ; thus

(50 -Pk)+ (q -y a ] ,

Now + lc+ l

the sum of all the letters a, b, l ;

g+p l

the sum o f the products taken two at a

time of all the letters a, b, .h

,l ;

the sum o f the products taken three at a timeo f all the letters a

,b,

l ;

u l = the product o f all the letters.

Hen ce,if the laws hold when n — l factors are multi

‘plied together, they hold when n factors ar e multipliedtogether ; but they have been shewn to hold when f ourfactors ar e mul tiplied together, th er ef0 1 e they he ld whenfive factors ar e multiplied together, and so o n : thus theyhold un iversally.

B INOM IAL THE OREM

We shall write the result for the mul tiplication o f n

factors thus for abbreviation(x + a)

—l+ Q$

n- g

+Ex " V.

NowP 1s the sum o f the letters a, b, c, . .

,h l

,which ar e

n in number , Q is the sum o f the products o f thesen (n 1 )1 2

of these

products ; and so

letters two and two, so that there ar e

n (n —l ) (n —2)products ; R IS the sum of1 2 0 3

on . See Ar t. 4 21.

Suppose b, l each equal to a . Then P becomesn (u — 1 ) 2

n (n — l ) (n —2 )a ,R becomes

and so on . Thus finallyn (n l )1

92(n —1>(7la‘V” _ 4

+1 . 2 3 4:

4 26 . Th e formula just obtain ed is cal led the Bin omialTheorem ; the series o n the right- hand s ide is called theexpan sion of (m+ a) and when we put this series insteado f we are said to exp and (x + Th e theoremwas discovered by Newton .

n a, Q becomes

(a: a)”

x”

n aat’” 1 a

3:c" —3

It will be seen that we have demon strated the theoremin the case in which the expon en t n is a p ositive in teger ;and that we have used in this demon stration the methodofma th ema tical induction .

4 27. Take for example (a'+ a)s. Here n 6 ,

n (u — 1 ) n (n — l ) (n —2)1 .

n (n—l ) (n —2) (n —3 ) 3_

1 . 2 . 3 . 4"

n (u— l ) (n —2) (n —3 ) (n —4 ) 6 . 5 . 4 .

1 . 2 . 3 .

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BINOM IAL THEOREM .

Hence, collecting the terms, we obtain (1 +2x- 1 8x +20x2+ 8x3—26x4 - 8x 5 + 20x6 - 8w7 + as.

4 29. I n th e exp an sion of (1 + x)n the coefiicien ts of

terms equa lly distan t f r om th e begin n ing and th e enda r e th e same.

Th e co efiicient o f the r01 term from the beginning

'

is

n (u l )(n —r + 2)by mul tiplying both numerator

and den ominator by ln r 1 this becomes

Th e r“ term from the end is the (n term fr om

the beginn ing, and its coefficient is

n (n —(n that is

by multiplying both numerator and den ominator by Ir 1

VJthis also becomes

4 30. Hitherto in speaking o f the expan sion ofwe have assumed that n den otes some ositive in teger .

But the B in omial Theorem is also app'

cd to expandwhen n is a positive fraction , o r a n egative quan

tity whole or fractional. Fo r a discussion o f the B in omialTheorem with any expon en t the studen t is referred to thelarger A lgebra ; it will however be a useful exerc ise toobtain various particular cases from the general formula.

Thus the studen t wil l assume for the presen t that whateverbe the val ues of a', a, and n ,

n (n l )(n 2)a, ‘n ae

“ 1 2a x c

“

n (n —1 ) (n—2) (n —3 ) 3a .i

°

If n is not a positive integer the series never ends.

BINOl ll l AL THEOREM .265

4 3 1 . As an example take Herc;in the formula

o f Ar t. 4 30 we put 1 fo r a', y fo r a, and

5fo r n .

n (u—l ) (n —2)

n (n 1) (n 2) (n 3 )

and so on . Thusis 1 1

21

3(H e) = l +éu

—§ y

+1—

6y

As an other example take (1 y) 5. Here we put 1 for a',

y fo r a , and—1 for n .

2

n (u—l ) 3 n (u 5

1 - 2 8’ 1 6

’

” (n — l l (n —2Xn —3 ) 3 5

128, and so on . Thus

” 2_ 1 2 2(1 +y) — 12g+

8y

Again ,expand ( l Here we put 1 fo r x, y fo r a,

and —m for n .

n (u 1 ) m (m 1 )

n (n —l ) (n —2)

n (n —l ) (n —2) (n —3 )1 2 3 4

266 EXAMPLES . XLII.

Thus ( 1 + y ” m— l —my 4.2)y”

3 )y'

As a particul ar case suppose m : 1 thus(1 —

y + 2/2 -

zl3+ e

4

This may be verified by dividing 1 by 1 y.

Again , expand (1 + 2x—a2)if in powers of 33. Put y for

2x x2; thus we have (1 Zak - at

?)é yfi

_ 1 11 +

2y sy

2+1 6y

1 —2x—a'2)s(2r a

“

)2 —

1

6a: (2w

Now expan d (2x (2x and collect theterms : thus we shall obtain

(1 + 2x —x2+ x3 —gx 4 +

EXAMPLES. XLII.

1 . Write down the first three and the last th ree termso f (a

2. Write down the expan sion o f (3

3 . Expand (1 - 2y)7

4 . Write down the first terms expansion(s q

5 . Expand (1 - l~ a'

6 . Expand (1 a:

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SCALES OF NOTA TION .

XLIII. Scales of Notation .

4 32. Th e studen t will o f course have learn ed fromA rithmetic that in the ordinary method o f expressingwhole numbers by figures , the number represented by eachfigure is always some mul tip le of so me p ower of ten . Thusin 5 23 the 5 represents 5 hundreds, that is 5 timesthe 2 represen ts 2 ten s, that is 2 times 101 ; and the 3 ,which represen ts 3 units, may be said to represen t 3 times

see A rt . 324 .

This mode of expressing whole numbers is cal led thecommon sca le of n otation , and ten is said to be - the baseo r r adix o f the common scale.

4 3 3 . We shall n ow shew that any positive in tege rgreater than un ity may be used in stead o f 10 fo r the radix ;and then explain h ow a given whole numbe r may beexpressed in any pr oposed scal e.Th e figures by mean s o f which a number is expressed

ar e called digits. When we speak in futur e o f an y r adix

we shall always mean that this radix is some positivein teger greater than un ity.

4 3 4 . To sh ew th a t any wh ole n umber may be exp r essed in terms of an y r adix .

Let N den ote the whole number, r the radix. Suppo sethat r " is the highest power o f r which is n o t greater thanN ; divide N by r

"

; let the quotien t be a,and the re

mainder P : thusN a r

" P.

Here, by supposition,a is less than r

,and P is less

than r". D ivide P by W“

; let the quotien t be I) , and th eremainder Q : thus

P br '“ 1 Q.

Proceed in this way un til th e , r emainder is less than r :

thus we find N expressed in the mann er s hewn by thefollowing iden tity,

N =ar " br "‘ l+ er

" ‘ z+ + kr +k.

SCALES OF NOTA TION . 269

Each o f the digits a , b, c, h,h is less th an r ; and

any o ne or more of them after the first may happen to bezero .

4 3 5 . To exp r ess a given wh ole n umber in any p r o

p osed scale.

By a given whole number we mean a whole numberexpressed in words, o r else expressed by digits in someassign ed scale . If n o scale is men tion ed the common scale

is to be understood.

Let N be the given whole number, r the radix o f thescale in which it is to be expressed. Suppose h

, b,a

the requ ired digits, n + 1 in number, beginn ing with thaton the right hand : then

N = ar" br "

‘ 1cr

n ‘ ” h r k.

D ivide N by r , and let M be the quotien t ; then it isobvious that M + h

, and that theremainder is k. Hen ce the first digit is found by thisrule : divide th e given n umber by th e p r op osed r adix,an d th e r emainder is th efir st of th e r equir ed digits.

Again , divide M by r ; then it is obvious that ther ema in der is h ; and thus the second o f the requireddigits is found.

By proceeding in this way we shall find in successionall the required digits.

4 3 6 . We shall n ow solve some examples .

Tran sform 32884 into the scale o f which th e radix isseven .

7 4 6 97 5

7 671 o

Thusso that .the number expressed in th e scale o f which theradix is seven is 1 64 605 .

SCALES OF N0 TA TION

Tran sform 74 194 in to the scale o f which the radix istwelve.

12 174 19412 15332 10

12 5 1 5

12 1 1

Thus 124 4 - 6 123 4- 1 1

In o rder to express the number m the scal e o f whichthe radix i s twelve m the usual mann er

,we require two

new symbols, on e for ten , and the other for eleven : we willuse t fo r the former, and e for the latter. Thus the numberexpressed in the scale of which the radix is twelve is3 6e2t.

Tran sform 64 5 032, which is expressed in the scale o fwhich the radix is n in e

,in to the scal e o f which the radix I S

eight.8 [ 64 5 032

72782 4 .

The division by eight is performed thus : First eight isn o t con tain ed in 6

,so we have to find how often eight is

con tain ed in 64 ; hei e 6 stands fo 1 six times n in e,th at is

fifty- four

,so that the question is how often is eight co n

tain ed in fifty eight, and the an swer is seven times withtwo ovei . Next we have to find how often eight is co n

tain ed in 25,that is how often eight is con tain ed in twen ty

three,an d the an swer is twice with seven over. Next we

have to find how often eight is con tain ed in 70, that is howoften eight is con tain ed in sixty- three

, and the an swer isseven times with seven over. Next we have to find howoften eight is con tain ed in 73 , that is how often eight iscon tain ed in sixty—six, an d the an swer is eight times withtwo over. Next we have to find how often eight is con

tain ed in 22, that is how often eight is con tain ed in twen ty,and the an swer is twice with four over. Thus 4 is the firstof the required digits.

We will indicate the remainder o f the process ; thestuden t should ca1 efullv w01k it for himself

,and then com

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XLIV. In ter est.

4 38 . Th e subject o f In terest is discussed in treatiseso n Arithm etic ; but by the aid o f Algebraical n otationthe rules can be presented in a. form easy to understandand to remember.

4 3 9. In ter est is mon ey paid fo r the use of mon ey.

Th e mon ey len t is called the Pr in cipal . The Amo un t atthe end o f a given time is the sum o f the Prin cipal and th eInterest at the end o f that time.

4 40. In terest is o f two kinds, simp le and comp ound.

When interest is charged o n the Principal alon e it is calledsimp le in terest ; but if the in terest as soon as i t becomesdue is added to the principal , and in terest charged o n thewhole

,it is called comp ound in terest.

4 4 1 . Th e r a te o f in terest is the mon ey paid for the useo f a certain sum for a certain time. In p r ac tice the sum isusually £ 100, and the time is o n e year ; and when we saythat the rate is £ 4 . 5 s. per cen t. we mean that £ 4 . 5 s.

,that

£ 45, is paid for the use o f £ 100 for on e year. In theo r yit is conven ien t, as we sh all see , to us e a symbol to denotethe interest o f o n e pound for o ne year.

4 42. To find th e amoun t of a given sum in any giventime a t simp le in ter est.

Let P be the number of pounds in the principal, n thenumber o f years, r the in terest o f o n e poun d for o n e year,expressed as a fraction o f a pound, M the number o fpounds in the amoun t. S ince r is the in terest o f o n e poundfo r o n e year, Pr is the interest o f P pounds for o n e year,and nPr is the in terest o fP pounds for n years ; therefore

443 . From the equation M =P (1 + n r ), if any three o fthe four quan tities III , P,

n,r are given , the fourth can be

INTERES T. 273

4 44 . To fin d th e amoun t of a given sum in any

given time a t comp ound in ter est.Let P be the number o f poun ds in the prin cipal

,n the

number o f years, r the interest o f o n e pound for one year,expressed as a fraction o f a poun d,M the n umber o f poun dsin the amoun t. Le t R den ote the amoun t o f o n e pound ino ne year ; so that R : 1 + r . Then PR is the amoun t o f Ppoun ds in o n e year . The amoun t o f PR pounds in o n e

year is BRR , o rPRz; which is therefore the amoun t o fP

pounds in two years. Similarly the amoun t o fPR 2 poun dsin one year isPR ZR

,o r PR 3

,which is therefore the amoun t

o fP poun ds in th r ee years.

Proceeding in this way we find that the amoun t o f Ppounds in n years is PR”

: that is

Th e in terest gain ed in n years is

PR " —P or P (R "l ) .

4 4 5 . Th e Pr esen t va lue o f an amount due at the endo f a given time is that sum which with its in terest for thegiven time will be equal to the amoun t . That is, the Pr in '

c ip a l is the presen t value o f the Am oun t ; see Ar t. 4 3 9 .

4 4 6. D iscoun t is an allowan ce made for the paymen to f a sum o fmon ey before it is due .

From the defin ition of p r esen t va lue i t follows that adebt is fairly d ischarged by paying the p r esen t va lue at

on ce : hence the discoun t is equal to the amoun t duedimin ished by its presen t value.

4 47. To fin d th e p r esen t va lue of a sum of mon ey dueat th e end of a given time

,an d th e discoun t.

Let P be the number of pounds in the presen t value, nthe n umber o f years, r the in terest o f o n e pound for o n eyear expressed as a fraction o f a pound

, M the number o fpounds in the sum due

,D the discoun t.

Let R z l + r .

n a .

EXAMPLES. XLIV.

A t simple in te restM =F(I+ n r ), by Art. 4 42 ;

therefore PM

D =M P1 + n r

At compo rmd in terestM : PR”

,by Ar t. 4 44

therefore P D =M —P

4 48. In practice it is very common to al low thein ter est o f a sum o f mon ey paid before it is due ins tead o f

the discoun t as here defin ed. Thus at simple in terest iastead of

1the payer would be allowed M n r for im

mediate paymen t.

EXAMPLES. XLIV.

1 . At what rate per cen t. will £ a produce the samein terest in on e year as £ b produces when the rate is £ 0per cent ?

2 . Shew that a sum o f mon ey at compoun d in terestbecomes greate1 at a given rate per cen t. f0 1 a.given n umbe ro f years than it does at twice that rate per cen t. fo r halfthat numbs1 of years.

3 . Find in h owmany years a sum o fmon ey will doubleitself at a given rate of simple in terest.

4 . Shew, by taking the first three terms o f th e Bin omial series for that at five pe r cen t. compo undinterest a sum ofmon ey will be more than doubled in fifteenyears.

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276 MISCELLANEOUS EXAMPLES .

12. If a = 1,b= 3 , and c = 5 , find the value o f

2a3 —c)—b2(2a

1 3 . Simplify (a +b)2 — b) —{a(2b—2)—(b

2

1 4 . D ivide2x 5 x 4y 4x 3y

25x

t’

ys 4y

5 by x3

xyz2y

3.

x4 2x 3 x

2 l

x4

x2 1

1 6 . Find the L .C.M. o f x2—9x — 10, x

2— 7x — 30,

17. Simplify

1 5 . Reduce to its lowest terms

3 5

x2— 9x —10

+x2— 7x—3 0 x

2+ 4x + 3'

x — 2 x + 1 5 x

18. Solve3

3 2 1fi

_

19. SOIVG — l ) —o )— 4 .

20. Two persons A and B own together 1 75 shares ina railway company. They agree to divide, and A takes 85shares

,while B takes 90 shares and pays £ 100 to A . Find

the value of a share.2 1 . Add together a + 2.v—y + 24b, 3 a

—4x—2y—s1b,

x + y—2a + 5 5 b;

and subtract the resul t from 3 a b 3x 2y .

22 . Find the value o f7

,J7ab(2c’

ab) (2a 3b} ,

when a z 3 , b and c : 2.

23 . S implify {x (x a) a(x a)} {x (x a) n(a

o 0 tv 1 5 010

v.

1o D (l

t

l4 W1 06

F8

“

6 3 q , an d ve1 1fy t 10

result by multiplication .

25 . Find the of x 4 3x3 10 and x4 32 3 + 2.

77ZlIISCELLANEO US EXAIlIPLES . 2

b a

b2— 4a2 b+ 2a 2a —b'

27. Find the L. Cr M . o f x2— 4

,4x3 —7x—2, and

4x2+ 7x — 2 .

26 . Simplify

28 . Solve6

29 . A man bought a suit of clothes fo r £ 4 . 7s . 6d.

The trowsers cost half as much again as the waistcoat, and

the coat half as much again as the trowsers and waistcoattogether. Find the price of each garmen t.

3 0. A farmer sells a certain number o f bushels o fwheat at 73 . 6d. per bushel and 200 bushels o f barley at

4 s. 6d. per bushel, and receives al together as much as if heh ad sold both wheat and barley at the rate o f 5 3 . 6d. perbushel . Howmuch wheat did he sell ?

3 1. If a = 1,b= 2

,c d= o, find the value of

a — b+ c ad — bc

a — b— c bd + ae

32. Multiply together x a, x b

,x a, and x b;

and divide the result by x2 x (a b) ab.

3 3 . D ivide 8x 5 —x‘—’y3 gy 5 by 2x y .

3 4 . Fin d the o f 4x (x2+ 10)— 25 x —62 and

x2—7x + 10.

12x‘3 — 1 5xy + 3y

2'

3 5 . Reduce to i ts lowest terms6x

3 6x2y + 2xyz 2y

"’

3 6 . Simplify

3 7. Solve9

M ISCELLANE OUS EXAMPLES.

2x— l x + 4 5x—l

9O

3 9. A can do a piece o f work in on e hour, B and 0’

each in two hours : h ow long would A ,B, and 0 take,

working together ?40. A having three times as much mon ey as B gave

two pounds to B ,and then he h ad twice as much as B

had. Howmuch h ad each at first "!4 1 . Add together 2x + 3y + 4z, x - 2y + 5z, and

7x—y +z.

42. Find th e sum,th e difference, and the product o f

3x2 4xy 4g2 and 4x2 2xy 3y

2.

4 3 . Simplify2a—3 (b—c) {a 2(b c)} 2{a 3 (b

4 4 . Find the G .C.M . o f

a‘ + 67x2+ 66 and 1 .

x4—1

xx + 1

xfi—l

4 6. Find the L.e.M. o f x2— 4 , xZ 5x 6 , and x

2—9 .

3x3—4x'3—x 1 4

6x3 l 1x2 l ox + 7

4 8. Solve 3 (x 1 ) 4 (x 2) x).

4 9. Solve .J(9 +4 x) 5 2,Jx .

5 0. How much tea at 3s. 9d. per lb. must be mixedwith 4 5 lbs. at 3s. 4d . per lb. that the mixture may beworth 3 s . 6d. per lb. ?

5 1 . Mul tiply 3 a2 ab b2 by a2 2ab 3b9, an d divide

the product by a + b.

5 2. Find the e . o f 2x (x 15 and

2x3—5x2—6x - l 15 .

3 8. Solve

4 5 . Simplify

4 7. Reduce to its lowest terms

5 3 . Simplify

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M ISCELLANE O US EXAMPLES .

6 8. Solvex

‘Z — x — 6 5 x

x + 1 2

70. A father’s age is double that o f h is son ; 10 yearsago the father

’

s age was three times that of h is son : findthe presen t age o f each.

71 . Fin d the value when x = 4 o f

J(2x + l )

69 . Solve

3x 3 1 6x2+ 23x 6

2x 3 — l 1x 2+ 17x — 6

and find its value when x 3 .

73 . Resolve into simple factors a” 3x 2,x’ 7x 10,

and x2 6x 5 .

72. Reduce to its lowest terms ;

1 3 474' S lmph fy

x2— 3x + 2

+x9 — 7x + 10 x

2—6x + 5'

175 . Solve x +

5 (4x

76 . Solve 9x 2 68 0.

77 A man and a boy being paid for certain days’ work ,

the man received 27 shillings and the boy wh o h ad beenabsen t 3 days o ut of the time received 12 shillings : h ad theman in stead o f the boy been absen t those 3 days theywoul dboth have claimed an equal sum . Fin d the wages o f eachper day.

78 . Extract the square root o f 9x 4 6 .v’+ 7.v

2- 2.v+ 1

and shew that the result is true when x : 10.

79. If a b 0 d,shew that

( tic -Fae2 b9d+ bd2 (a + c)a

80. If a,b,c,d be in geometrical progr ession , sh ow that

a,2 d2 is greater than b2 c

”.

b8 1 . If n is awh o le positive number + 1 is divisible

y 8 .

M ISCELLANEOUS EXAMPLES .281

82. Find the least common multiple o f x2 4y

2,

ai + 6x9y + 12xy

2+ 837

3,and x

3 6x3 y + 12xy2— 8y

3

83 . Solvex y 2 x y

84 . Solve x 2+ 2x 2 ,J(332+ 2x 1) 4 7.

85 . The sum o f a certain number con sisting o f twodigits and of the number formed by reversing the digits is1 21 and the product o f the digits is 28 find the number.

86 . Nin e gallons ar e drawn from a cask full o f win e,

and it is then filled up with water then n in e gall on s o f th emixture are drawn ,

and the cask is again filled up withwater. If the quan tity o f win e n ow in the cask be to thequantity of water in it as 16 is to 9, find how much the caskholds.

87. Extract the square root o f1 6x

6 25g6 3 0xy

’5 24 .1:4

y2

4 ox3ya.

88. In an arithmetical progression the first term 13 81 ,and the fourteen th is 1 5 9. In a geometrical progressionthe second term is 8 1 , and the sixth is 1 6 . Find theharmon ic mean between the fourth terms o f the two progressien s.

89. If find the value to five places o f

6dec1mals o f

J5 1

90. If x be greater than 9,shew that 4 x is greater

than

9 1 D ivide (x y)3 2y (x y )

2

y2(x y) by (x 2y)

2

92. Find the G . c. M . and the L. o f

24 (x3 +x2y + xy

2+ 31

3) and 16 (x

3—x y + xyz—y3).

9 3 . Simplify7 l 1x J

x3 + x3

y + xy + y3+x3 x

2

y + xy—y"

x2y2

x2+ y

2°

M ISCELLANEOUS EXAMPLES.

6x + 7+2x + 5

94 . Solve1 3

\ 95 . Solvexy + 20 (x yz+ 30(y 3x—2z= 0.

96. Solve 3x2—2x + N/(3x2—4x 18 + 2x .

97. A rows at the rate o f 8%miles an hour. He leavesCambridge at the same time that B leaves Ely. A spends12 minutes in Ely and is back in Cambridge 2 hours and

20 minutes after B gets there. B rows at the rate o f 745miles an hour ; and there is no stream. Find the distan cefrom Cambridge to Ely.

“

9 8. An apple woman finding that ap les have thisyear become so much cheaper that sh e co d sell 60 morethan sh e us ed to do for five shill ings, lowered h

’er price andsold them on e penny per dozen cheaper. Find th e priceper dozen .

99. Sum to 8 terms and to infin ity 13,100. Find three numbers in geometr ical progression

such that if 1 , 3 , and 9 be subtrac ted from them in orderth eywill form an arithmetical progression whose sum is 1 5 .

101 . Multiplyxt’f —xS -t xg—z uxi—mm é- 1 by xh l ;

and divide l —xg by 1l.

102. Find the L. 0.M . o f xS—a 3

, x3 +as

,

xs—axJ and x3 + ax ’—d

’x—a’.

a3—b3

104 . Solvex + 5 1 4x—14 x + 10“

(i 2+ (13 4- 258 )

10

105 . Solve — + i7 18

x - l x—5 x + l+x + 5

'

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i

284 M ISCELLANEOUS EXAMPLES .

\ 1 18 . Simplify1 19 . Th e third term o f an arithmetical progression is

18 ; and the seven th term is 30 : find the sum of 17 terms.

120. If6

30be in harmon ica] progressi on ,

shew that a,b, c ar e in geometrical progression .

11 21 . Simplify a

122. Extract the square root o f3 7xzy

2 3 ox3y 9x 4 2Oxy3

123 . Resolve 3x3 14x2 24x into its simple factors .

x + 5 3 (5x + 1) 4124' some

2x — l 5 x + 4 2x — l_

Solve x34x

126 . Solve x2—y2= 9, x + 4 = 3 (y

127 . Solve y + J(x2 J(x + 1) - J(x

128 . If a , b, c, d ar e in Geometrical Progression,a b+ d 0

3o2d+ d3.

129. Th e common differen ce in an arithmetical progre ssion is equal to 2, and the number o f terms is equal tothe secon d term : find what the first term must be that thesum may be 3 5 .

130. Sum to n terms the series whose mth term is

2 x 3"

.

1 J(1—2.x)1 N/(1—2x )

1 32. Find the M . o f 30x ‘ 16x3 5 0x? 24x and

24x“ 3 2x .

1 3 1 . S implify

M ISCELLANEOUS EXAMPLES .28 5

1 3 3 . Solve x2—x

1 3 4 . Form a quadratic equation whose roots shall be3 and —2.

a; 1 3 5 . Solve x4+—1—= a4 +

13 6. Solve

1 3 7. Having given find the value of6

J3 —1to five places o f decimals .

1 38. Extract the square root of 6 1 28 ,J3 .

1 3 9. Find the mean proportional between

140. If a,b, c be the first, second and last terms of an

arithmetical progression , find the number o f terms. Alsofind the sum o f the terms.

14 1 . If cl,0,b, a are 2, 3 , 4 , 5 , find the values o f

1 42. In the product o f 1 4x + 7x2 10x 3 1 5x“ by1 5 x 9x2 17x 43 find the co efiicien t o f x 4

D ivide 21x 5 2x 4 23x2+ 3 3x 27 by 7x2 4x 9.

a4

_ b41 43 . Simplify

a2+b2+ 2ab a

2+ ab

’

—1

~/x Jx + Ja x + a

M ISCELLANEOUS EXAMPLES.

Solve the followingGO—x 3x 5

1 4 7

x 4 5 x 12

x 3 4 3x

T

3x 5 y 5 x 3y

20 8

145 . Solve the following equation s :

8—x+6—x

J32 + 1 = 7 .

(3 ) 3x2—4xy z 7, 3xy—4y

2= 5 .

1 4 6 . A bill o f £ 20 is paid in sovereigns and crown s,and 3 2 pieces are used : find h owmany there were o f each

1 47. A herd cost £ 180, but on 2 oxen being stolen,the

rest average £ 1 a head more than at first : find the numbero f oxen.

148. Find two numbers when their sum is 40, and thesum o f their reciprocal s is£3

14 9. Find a mean proportional to 2§ and 5 5 ; and a

third proportional to 100 an d 13 0.

\1 5 0. If 8 gold coin s and 9 silver coins are worth as

much as 6 gold coin s and 1 9 silver on es, find the ratio o fthe value of a gold coin to that o f a silver com.

1 5 1. Remove the brackets from(x—a) (x—b) (x—c) [bc (x —a)

1 5 2. Multiply a 2 (a?b) 2 ,Jb

’

by a 2U(a2b) 2Jb.

15 3 . Find the O . C. H . o f x4—16x3 + 93x’—23 4 x+ 216

and 4x3—48x’+ 186x 23 4 .

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M ISOELLANEO US EXAMPLES.

162. Find the o f 18a3 18a2x Gax2 6x3,and

6oe z 75 d x 1 5x2.

1 6 3 . Find the L. C.M . o f 18 (x24 (xi y

3).

164. Solve the followin g equation s :2x 4 3x 2

7 5

9x + 2o — 12 x

3 6 5 x — 4 4'

3

2 (x —4y), 14 (x + y )

16 5 . Solve the following equations

(1) 3 212— 5 232 : 12.

N/(2x 3 ) N/(x — 2) 1 5 .

X03 ) x

2+ y2= 290, xy 14 3 .

3x2—4y r: 8, 5 x2 6xy = 32.

1 66 . A and B together complete a workwhich woul d have occupied A alon e 4 days :would it employ B alon e ?

9fax Fin d two numbers whose product is 5 of the sumof their squares, and the differen ce of their squares is

96 times the quotien t o f the less number divided by th egreater.

1 68. Find a fraction which becomes3

1

;o n in creasing its

numerator by 1 , and 21

;on similarly increasing its den omi

n ator.

M ISCELLANE o US EXAMPLES . 289

1 69. If a b c d, shew that1 I

.

a b a b c d c d

170. Find a mean proportion al between 169 and 25 6,and a th ird proportional to 25 and 100.

1 71 . Remove the brackets from the expressionb—2 {b—3 [a—4 (a

172. Simplify the following expressionsx 2x2 y

2 3xy2 3x3 y

3 4xy3 2xzy

2y‘

y xy xzzr

(p —q—m)p - (m +q

1 73 . Find the G. o f x4

ax" 9azx2 1 l a3x 4 a4

and x4

d x3 3em2 5 a3x 2a“.

174. Solve the following equations2x + 1 . x + 7

3 5

l ox + 1 7 12x + 2 5 x—4

18 1 3x—16

(3 ) 9x 70, 7y

6x + 7 2x +~ 1 9

3x + 1 x + 7

1 75 . Solve the following equations7x - 8

x

(2) 2x2— 3y22 2

, xy z 20.

(3 ) 2y2 3x2—4xy z 7.

(0 +y = 6, x3 +y

3= 126.

( l ) x

(1) x + 4

M ISCELLANEOUS EXAMPLES.

1 76. When are th e clock- hands at right angles firstafter 12 o ’clock ?

177. A number divided by the product o f its digits1

"

g1ves as quotien t 2, and the digits are inverted by adding27 : find the number.

1 78. A bill of £ 26 . 1 5 3 . was paid with half-guin eas andcrown s, and the number o f half- guin eas exceeded the number o f crown s by 17 : find h owman y there were o f each.

1 79. Sum to six terms and to infin ity 12 + 8 5%

180. Extract the square root o f 5 5 —7

181. If x — N/“ lfind the value o f

3x2—16x—12

183 . If two numbers o f two digits be expressed by th esame digits in a reversed order, shew that the difference ofthe numbers can be divided by 9.

184 . Solve the following equations3x — 3 3x—4 2 1

4 3

1 4 —x_

(3 ) Mx + 1

185 Solve the following equation s(1) J(x + 3) x ,J (3x

(2) 2) J(3 .2 4 ) 8 .

(3 ) x4—x2 (2x—3 )= 2x + 8 .

186 . Find two numbers in th e proportion o f

such that the s uare o f their sum shall be equalcube o f their di erence .

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M ISCELLANEOUS EXAMPLES .

195 . Solve the following equations3 l

(1) yer —3 )

803

(2)

(3 ) x + y = 6 ,

196 . Th e express train between Lon don and Cam

bridge, which travels at the rate o f 3 2 miles an hour, performs the j ourn ey in 25 hours less than thetrain which travels at the rate o f 1 4 milesthe distance.

197. Find the number, consisting o f two digits, whichis equal to three times the product o f those digits , and isalso such that if it be divided by th e sum o f th e digits thequotien t is 4 .

198. Th e number o f r esiden t members o f a certaincollege in the M ichaelmas Term 1864 , exceeded the number in 1863 by 9 . If there had been accommodation in1864 fo r 1 3 more studen ts in college rooms , the n umber incollege would have been 18 times the number in lodgings,and the n umber in lodgings would have been less by 27than the total number o f residen ts in 1863 . Ifi

'

nd thenumber o f residents in 1864 .

1 99. Extract the square root o fa4 2a3b 3a2b2 2ab8 b“,

and o f (a b)‘ 2 (a

2 b?) (a b)2+ 2 (a4+

200. Find a geometrical progression of four termssuch that the third term is greater by 2 than the sum o f

the first and second, and the fourth term is greate r by 4than the sum o f the second and third.

719—5 5 4- 3 03:

6 —3x

202. Find the G.C.M . ofx‘+ 4x

9+ 16 andx ‘—x’ + 8x - 8.

9—2x

MISCELLANEOUS EXAMPLES 293

1 23 ;—5 x2—x + 6

203 Add together2 + 3x

’

(2+ 3x“

1

1—x + x r

Solve the following equation s3x 5 21 x

8 33 9— 5x .

(2) —x).

2x 331 g1 6 12

4 73 2” 3 ‘

205 . Solve the following equations :

(1) 6x +3 5

x

(2) 4 (x2+ 3x)

(3) x2+ xy = 1 5 , y

2+ xy = 10.

206 . A person walked out from Cambridge to a villageat the rate o f 4 miles an hour, and on reaching the railwaystation had to wait ten m inutes for the train which wasthen 4%miles ofi

'

. O n arriving at his rooms which werea mile from the Cambridge station he found that he h adbeen out 34 hours. F in d the distan ce of the village .

207. The ten s digit of a number is less by 2 than theunits digit, and if the digits are inverted the n ew numberis to the former as 7 is to 4 : find the number.

208. A sum o f mon ey con sists o f shill ings and crown s,

and is such, that the squar e o f the number of crowns isequal to twice the number o f shil lings ; also the sum isworth as many fio r ins as there ar e pieces of money : findthe sum.

209. Extract the square root o f4x 4 8ax3 16b3x2 16ab2x 1 664 .

2 10. Find the arithm etical progression o f which th efirst . te rm is 7, and the sum o f twelve terms is 3 48.

M ISCELLANEOUS EXAMPLES .

211 . D ivide 6x5 25x‘y

—4 9x2y3 62xy

4

by 2x2 7xy 9y

2

212. Multiply12 + 4 1x + 3 6x2

b 5 —2x +26x — 8x3—14

4 + 7xy

3—4x3 + 5x

213 . Reduce to its lowest terms4x3 - 4 5 x2+ 162x 185

x4 1 5x3 + 81x

2 l 85 x + 15 0°

214 . Solve the following equations3x—2 1—5 x

5 11

2 2 — 17 +1x— 8x +

4y y 74

1 1 1 1 1 4 1 1 5(3 ) 2 ’

x+x 9

’

y+z 18

215 . Solve the following equation s :1 l l

x x + 3 6°

(2) 10xy 7x2 : 5 y‘3—3xy z 20.

(3 ) x + y - 6, x4+ y

4z 272.

216 . D ivide £ 34 . in to two parts such that the number of crowns in the on e may be equal to the number ofshil lings in the other.

2 17. A number,con sisting of three digits whose sum is

9 , is equal to 4 2 times the sum o f the middl e and left -handdigits ; also the right- hand digit is twice the sum o f theo ther two : find the number.

218. A person bought a nmnber of railway shares whenthey were at a certain price for £ 2625 , and afterwardswhen the price o f each share was doubled, sold them allbut five for £ 4000 : find how many shares he bought.

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296 M ISCELLANEOUS EXAMPLES.

a daughter, and th e widow £ 5 00more than all the five children together : find h owmuch each person obtain ed.

228. A cistern can be filled by two pipes in 12; hours.

Th e larger pipeby itself will fill the c istern soo n er than

the smaller y 2 hours. Find what time each will separately take to fill it.

229 . The third term o f an arithmetical progression isfour times the first term ; and the sixth term is 17 : findthe series.

230. Sum to n terms 3 § 1325 +

23 1. Simplify the following expression sxa -l- b-H

X xa -l- b—c

X xa - b-l- c

X xb a + b

+( 52+ b2

a +b 2a 2a (a—b)’

— ab+ b2 £12—b?

x2+ 1 1x + 30

232. Reduce to its lowest terms9333 + 5 3x2 9x 18

23 3 . Solve the following equations1 1 1 7

(I)x+2x 3x

'

3°

—x -

y ,

234 . Solve the following equations1 65

x + 3 x + 10

(2) —c)x Ja .

x9+ y

z= 4 h

MISUELLANEOUS EXAMPLES .

‘ 297

235 . A body of tr00ps retreating before the en emy,from which it is at a certain time 26 miles distan t, marches18 miles a day. The en emy pursues it at the rate o f 23

miles a day, but is first a day later in starting, then aftertwo days’ march/

is forced to halt for on e day to repair abridge, and this they have to do ag ain after two days’ moremarching. After how many days from the beginn ing of theretreat will the retreating force be overtaken ?

236 . A man has a sum o f mon ey amoun ting to £ 23 1 5 5 .

cons isting only o f half- crown s an d fio r in s ; in all he h as 200pieces o f mon ey : h owmany has he o f each sort 2

237. Two numbers are in the ratio o f 4 to 5 ; i f o n e isin creased, and the other dimin ished by 10, the ratio o f theresulting numbers is inverted : find the numbers.

238. A colon el wished to form a solid square o f h ismen . The first time he had 3 9 men over ; the second timebe in creased the side o f the square by on e man , and thenhe foun d he wanted 5 0 men to complete it. O f how manymen did the regimen t con sist ?

239. Extract the square root o fa6 2a5b 3 2 462 4a3b3 3 2 264 2265 be,

and of a2+ 4bz+ 902+ 4ab+ 6ac + 1260.

240. Multiply xgyé 2xy 4xéy3 by x2

24 1 . Simplify4 0x31 (9x Sy) (5 x 2y) (4y 3x) (1 5 x

1 + x l —x l — x —i- x 2 1 +x + x2

1 —x 1 + x l - l—x z

242. Find the G. o f x‘ax

” 2a2x2 Sa3x a‘,

and x4

ax 3 3a3x ab‘zx a

2b2

243 . Two shopkeepers wen t to the cheese fair with thesame sum o f mon ey. The on e Spen t all h is mon ey but 5 3 .

in buying cheese, o f which he bought 25 0 lbs. Th e other

M ISCELLANEOUS EXAMPLES.

bought at the same price 3 5 0 1bs ., but was obliged toborrow 3 5 s. to complete the payment . How much hadthey at first i

The two digits o f a number are inverted ; thenumber thus formed is subtracted from the first, and

leaves a remainder equal to the sum o f the digits ; the difference o f the digits is un ity : find the number.

245 . Find three numbers the third o f which exceedsth e first by 5 , such that the product o f their sum multiplied by the first is 4 8, and the product o f their sum multiplied by the third is 128.

24 6 . A person lends £ 1024 at a certain rate o f

in terest ; at the end o f two years he receives back for h iscapital and compound in terest on it the sum o f £ 1 1 5 6 :find the rate o f in terest.

247. From a sum o f mon ey I take away £ 5 0 morethan the half

,then from the remainder £ 30 more than the

fifth,then from the second remainder £ 20 more than the

fourth part ; at last on ly £ 10 remain s : find the originalsum.

248. Find such a fraction that when 2 is added to thenumerator its value becomes 5, and when 1 is taken fromthe den ominator its val ue becomes 1

249. If I divide the smaller o f two numbers by thegreater, the quotien t is '

21, an d the remainder is ‘

04 1 62 i fI divide the greater number by the smaller the quotien t is4, and the remainder is ‘

742 : find the numbers.

(ivy-F (will )2s a?

25 0. Shew thatx y

25 1. Simplify6a [4a {8b (2x 4b)—22b} 7b]

—[7b {8a (36 4a) 8b} be ].

25 2. Mul tiply a x successively by a x , a2

x3,a‘

x‘

,

also multiply am‘ ” b’“ P by b? “

c.

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M ISCELLANEOUS EXAMPLES.

262. Multiply by x2—l ; and

x 2a 2x_ l by

a

x a a

263 . What quan tity, when mul tiplied by x

give x3

264. Simplify the following expressions3x3 1 3x2+ 23x - 21

6x3 + x 2—4 4x + 21

a + b a — b

2W—w 2m+® fi

265 . Solve the following equations5 x + 3 2x—3

x — l 2x—1

<2) ..lx

(3) + 9x = 167.

266 . Solve the followin g equation s :(1) x

g—x —Gz o.

x + 1+x + 2 2x + 1 3

x — l x —2 x + 1

(3 ) x2

x + y z 5 .

267. The ratio o f the sum to the difference o f twonumbers is that o f 7 to 3 . Shew that if half the less beadded to the greater, and half the greater to the less, t heratio o f the numbers so formed wi ll be that o f 4 to 3 .

268. Th e price o f barley per quarte r is 1 5 shillingsless than that o f wheat, and th e value o f 5 0 quarters o f

barley exceeds that of 30 quarters o f wheat by £ 7.find the price per quarter o f each.

M ISCELLANEOUS EXAMPLES 301

269. Shew that

(bcd+ cda d'

ab abc)2 (a b c d)2abcd

(be ad) (ca bd) (ab ed).

270. Extract the square root o f5 x2 x 13x

4+x

1 2

an d of 3 3—20J2.

271. If a = y +z—2x , b= z+ x —2y , and c=x +y —2z,find the value of bz+ cz+ 2bc—a2.

272. D ivide x4—21x -1- 8 by 1—3x +x2.

a x a x a?

x2

273 . Add togetherd _ x

,x +x a

Z—xz'

27az 3xx 7x 2

3a —x 1 5 xz+ ar.x—2x2

12ax 5 xZ

D in de 11 —x2

275 . Simplify 1a +

276. S olve the foll owing equation s :

JIISCELLANEOUS EXAMPLES.

277. Solve the following equation s

(1) e2(x a)

2= b2 (x

x

+5 x + l

x 2 x 3

(3 ) J(13x—1 ) - J (2x

278 . A person walked to the t0p of a moun tain at therate o f 2 miles an hour, an d down the same way at therate o f 3 miles an hour, and was out 5 hours : h ow far

fid he walk altogether i

279 . Shew that the difference between the square o f anumber, con sisting o f two digits, and the square o f thenumber formed by changing the places o f the digits is divis ible by 99.

280. If a b c d, shew that

,J(a2 b?) me

? d?) ,3/(a3 63) d3).

28 1. Fmd the value of“NZ

, b?)when a = 3 , b= 4 .

282. Subtract (b a) (c d) from (a b) (c what isthe value o f the result when a : 2b

,and d : 20 ?

283 . Reduce to their simplest formsx2—2ax —24a.2 x—y x

.fri‘— ’

7ax —4 4 a2 x + y x—y y—x

284. Solve the equation s :4 1 9

3 + x x 7x'

3x — 2y x—y

5 2

(3 ) J§2x 1) N/(3x + J (1 1x +

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M ISOELLANEO US EXAMPLES.

d)2 (ad be)2

02 d 2

bc (b2 a

2) + ab (a~ b?)

c) + b2(c 0.

9 92. Simplify (ao—b

293 . I f shew that a 3 + b3 +fi z 3aba

294. Reduce to its lowest termsx4+ 2x3 + 6x —9

x4+ 4x3 + 4x

2— 9'

295 . Solve the following equation s10x + 1 7 l 2x + 2 5 x — 4

1 8 l 3x—1 6 9

(2) 6x—5 y = 1, y—l z.

(3 ) + 8x = 129.

296 Solve the following equationsx l 3x 1

4 x 4

(2) N/(2x 2) N/(4x 3 )= 20.

(3) ,J(3x + 1)—J (2x

297. A siphon would empty a cistern in 48 minutes,a cock would fill it in 3 6 minutes ; when it is emp yt bothbegin to act . find how soon the c istern wil l be fi lle

pd.

y

A298. A waterman rows 30 miles and back m 12 hours,

and he finds that he can row 5 m iles with the stream in

the same time as 3 again st it. Find the times of rowingup and down .

299. In sert three Arithmetical means between a—band a + b.

\3 00. Find x if 21 2 $2“ 8 1 .

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308 ANSWERS.

a + x . 3 5 . a3+ b3. 3 6. x

2- dx -i- a

2.

3 8 .

(x—5 ) (x 4 0. (x 4 1 . (x

(x 4 ) (x 1 4 3 . (x 3 ) (x 3 ) (x2+

(x 5 ) (x2 5 x

(x—2) (x 2) (x2+ 4 ) (x

4

(x (x2

(a 4b) (a 5 b). 4 8 . (x—Gy) (x 7y).

(a + b—5 c) (a +b

(2x + 2y—a —b) (x + y 3a 3b).

XII. 1 . 3x2

2 . 4a2b’.

4 . 7a2b3x3y

3. 5 . 2 (x l ). 6 . 3 (x

7. 8 . x2—y2. 9. x + 5 . 10. x—7.

l l . x—l o. 12 . x—12. 1 3 . x2+ 3x + 4 .

14 . x2— 5 x + 3 . 1 5 . x

2- 6x + 7. 1 6 . x

2—6x — 5 .

1 7. x + 3 . 18. x — 4 . 1 9. x2—x + 1 .

20. x2—x + 1. 2 1 . 3x + 2. 22. x

2—x—1 .

23 . 22 4 2. 24 . .2 — 2. 25 . x

2+ 1 .

26 . x2+ 3x + 5 . 27. 7x

2+ 8x + 1 .

28. x4—2x3 - l- 3x

2—2x + 1 . 29. x2

30. x + 1 . 3 1. x + 7. 32. x + 3y . 3 3 . x +c .

3 4 . x—2a . 3 5 . x—y .

XIII. 1 . l 2a2b2 2 . 3 6a 3b2c3. 3

,24a

2b2x3y3.

4 . (a b) (a b)2. 5 . l 2ab (a

3 b3). 6 . (a b) (a3 b3) .

7. 8 .

9. x (2x + 1) (3x 1) (4 33

+

10. (22 4 5 + 6) (x 1) (517

1 1. (x2 3 8 + 2) (x 3 ) (x +

(x2+x + 1) (x

2+ 1 ) (x + 1) (x

13 . (x3—x

2- 1) (x

14 . (x2

ax (12

) (x2

ax a2

) (x —ar.)2

3 6a2b362. 16. 120 (a b)2

(a b)2

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ANSWERS . 3 13

ab—p q 15

a + b+p + q5 5

2(x +b - l- 3) 6

1

1 360. 5 0 6 1 . 25 . 63 . (a—b)

2. 64 . a.

8 1°

XXI. 1 . 30. 2 . 2. 3 . 4 . 70.

5 . 1 7, 3 1 . 6 . 7. 28. 8 . November 2oth .

9 . 5 2. 10. 1 1 . 12. 14 , 24 , 3 8.

13 . 14 . 103 . 1 5 .

17. 18 . 10. 1 9. 20.

21 . 22. 14 . 23 . 24 . 21 .

25 . 26. 24, 60, 192. 27. 840. 28. 3 0000.

29. 4 20. 3 0. 24 . 3 1 . 5 00. 32 .

3 3 . 3 4 . 5 0, 3 5 .

3 6. 24,

3 7. 88, 44 . 3 8.

3 9. 1 3,27. 4 0. 4 1 . 8 5 , 3 5 . 4 2. 1000.

1 8, 4 4 . 24000. 4 5 . 80. 4 6. 26, 1 6, 3 2, 27, 42.

4 8. 104203.

1 . 72. 2. 20, 30. 3 . 200 miles from

Edinburgh. 3 4 . 12,16 . 5 . 6 .

7. 4 8 . 8. 9. 10. 3 0. 1 1 .

12. 10, 60. 14 . 10 shillings. 1 5 . 5 5 , 4 5 .

1 6 . At the end o f 5 6 hours. 1 7. 27, 1 7. 18. 168, 84 , 42 .

19. 20. 14 4 days. 2 1. 1 5 , 21 .

22. 25 60. 23 . 3 6, 5 4 . 24 . 60 25 . 12. 26 . 8 pen ce.27. 875 , 1 125 . 28 . 25 . 29. 10

,20. 30. 20

, 80.

3 1 . 32. 4 0, 5 0. 3 3 . 3 4 . 28.

3 5 . 24 . 3 6 . 1024 3 7. 4 5 0, 270. 3 8. 2200, 1 620,

1 100,1080. 3 9 . 60. 4 0. 4 1 . 3 0.

42. 6 0. 4 3 . 240. 4 4 . 3d. 9d. 1 3 ./i d. 4 5 . 5 0d .

4 6 . £ 13 34. 4 7. 24 . 4 8. 60. 4 9. 41120000.

5 0. 25 . 5 1. 5 2. 3 9. 5 3 . 40.

3 141 ANSWERS .

5 4 . 200000000. 5 5 . 68 . 5 6. 4 8. 5 7. 49711“ minutes

past three. 5 8. 3 23

3

1minutes past three. 5 9. £ 288.

60. 2 seconds. 6 1 . 40 minutes past eleven.

62 . £ 300 and £ 200. 6 3 . 1 4. 64 640.

XXIII. 1 . 10 ; 7.

4 . 4 ; 1 . 5 5 5

8. 2 ;— 3 .

12. 10 ; 73 8 1 5 216 ‘ 1 5 a 5 7

'

20. 1 3 ; 5 .

24 . 5 ; 7

28. 12 ; 3 .

3 2. 2 3

3 7. b ; a.

4 1 . a ; b.

2 . 2c c

42. a + b,a b. 4 3 . b) . 4 4.

a + b’

a + b'

XXIV. l . 2 .

- 3 .

4 . 5 . 4 ; O ; 5 . 6 . 5 ;

7. 4 5 ; 8 .

2 3 2 1

3,4

,5

. 1 1 . a —a), &c.

12. x =§ - a,&c. 1 3 . x &c.

abc14 . 1 5 . x — a

, y— b, z— c.

1 6. x = 4, y : 5

,

XXV. 1 . 2 . 3 .

4 . 24 ; 60. 5 . 30714 871. 6 .

1

4

5.

8 . 4 5 ; 63 . 9 . 72 ; 60. 10. 30d . ; l sd. 1 1 . 5s . ; 3s.

3 . 2 ; 13 .

7. 20 ; 10.

“

l l . 3 5 ; 4 .

1 5 . 6 ; 1 2 .

19. 20 ; 20.

23 . 4 ; 9 .

27 . 10 ; 8 .

3 1 . 3 ; 2.

3 5 . a ; b.

We a2bc

a2+ b2 a

2+ b2

’

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3 16 ANSWERS .

5 3 . 5 4 . 0. 5 5 . 13 5 6 . 6 ,- 34.

5 7. 5 ,—1 5 8. 5 9. 60. 2g, 0.

b(a + b)262. (a i b) . 6 3 . 64 . a,

2a + b°

XXVII. 1 . 2. 4 9 3 . 4.

5 . 5 ,—3 . 6 . 3 ,

—2.

9. 9,- 12. 10.

4 3 .

13 . 14. 14 . 1 6. 1 5 . 1 . 17. 4 .

4 b?) 318. 4 . 21 . 3a .

22. 0, 4$5 . 23 . 0, 4 5 . 24 . 0, 4 . 9 2. 25 . 2, 4 1 .

26. O,

27. ( 7,—2a

,—2a . 28. a ,

XXVIII. 1 . 2.

4 . 18, 12, 9 . 5 . 12,10. 6 . 4

, 6 .

8 . 9 . 1 1 . 10..7. 1 1 .

13 . 24 . 1 4 . 27 lbs . 1 5 . 83 . 9d. , 76 .

1 7. 18 . 8d. 1 9. 10, 9 miles

21 . 22. 9 gallons. 23 . 64 .

25 . 4 per cen t.

XXIX. l . 5 ,—4 ; 4 , —5 .

3 . 4 8 ° $ 6. 4 .- 4

7.—24

10. 6, 0 ; 5 , 0.

12 . 0. 1 5 . a, b.

4 . $4 .

7. 8. 12,

- 3 .

1 1 . 2, 1 2.

3 4

5’5

°

a—l

2

3 .

7. 196 .

12. 1 5 .

1 6. £ 20.

3 . 20. 5 6 .

24 . Equal .

5 0 7,_ 4 . 4

,—70

3 18 ANSWERS .

XXX. 1 . 2. 6 ; 18 . 3 . 8 ; 24. 4 . 8 ; 16 .

6 . 7. 7 ; 5 .

9 . 10. 1 1 . 12.

1 3 . 1 4 . 1 5 . 16 .

1 7. 18. 19 . 1 60 ; £ 2. 20. 24 ;

21 . 75 6 ; 3 6 ; 27. 22 . 4 ; walking ; 4 ; r owing at‘

first.

2 3 . 10 12 miles per hour. 24 . 6 miles.

XXX1. 1 . 8x3y9z12

. 2. 8x6y6z9 3 . Sl a ‘bsc" .

4 95 4 x“

27316

y°z3

7 . 07+ 7a“b 2151

5 23 5 a4b3 3 5 41364 2l a ’b5 7abc V.

9 . a“ 3a“b2 4 3a

2b“ b“. 10. 1 3x 3x2 x”.

1 1 . 1 2. 27 5 4x + 3 6x’ 8x3 .

1 3 . 1 4 . x4 8x3 + 24x

2- 16 .

1 5 . 16 . 2a 3x3 + 6axb2y2.

17. 2a4fi + 12a2x2b9y2+ 264y

4 18 .

19. 1 - 4x2+ 6x4—4xs+ x8. 20.

22. l + 2x—x2 23 .

24 . 1 —6x + l 5x2—18x 3 + 9x 4 . 25 .

26 .

28 . 1 3 16 5 —336.

29. 1 916 8x“.

3 0. l —9x + 3 6x2—8 1x 3 + 108x 4

3 1 . 3 2. l —2x + 3x‘

3 3 . 1 4x

3 4 . 4 (ab ad be 3 5 . 2 (a2 2ac c? 62 d

36 .

1 —12x + 60x2—1 601 3 + 240x4

3 8. 1 8x 70x4 8x7 £63.

3 9. 1—3x3 + 3x6—a:9 .

XXXII. l . 3a9b°. 3 .- 4ab’. 4 . 26117263 .

5 ab 3a aab 63 . 6 .

703 bc 2b“

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320 ANSWERS .

22. a: y . 23 . a 24 , b?!

03.

1 2 .

1 i 1 4 1 4 1

25 . x2+ 2x a

fi+ 3x a + 2x

8a + ca

? 26. x —2x 8

y H A.5

_

27. x5—2x

‘ i.

28. x — 2—x- 1

. 29. x°

30. 2x§ f s + 4x 3.

XXXIV. 1 . N 2. 2.

3 .

6 .

5

3”

7. 2 + 2J2—2J3 . 8 . 2+ -224—J I5

10. 5 + 2,J6 .

3 3

8 + 9 J6 + 4 J 1 5 + 6 J1 1 3 .

5 3

7 5 2 3 8

12’8

’3’4

’9

°

6 .

10.

13 . 14 .

XXXVI. l . 14 . 2. 18 4 . 12. 5 . 4 .

6 . 4 . 7. 8 . 5 . 13 . 4 5, 60, 80.

14 . 4 , 6, 9.

XXXVII. 1 . 4 . 3 .

6 . 5 . 8. abc. 10.

1 1. 15 . 12. £ 15 3 60.

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3 22 AM 8WERS .

15 . 7 + 1 .

4 s 1é(3x)

—sys.

13 1 6 1 ,

17. a.

'

b+ 65 a —m

a 8 63 +

11 9 ‘

as

XLIII. 1 . 2042132 . 2 . 22600. 3 . 1 1 101001010.

4 . 2076 . 5 . t4 5 92. 6 . Radix 8 . 7. Radix G

8 . 9 5 21 ; $6 . 9 . Radix 5 . 10. 6 66 .

XLIV. 1 . 3 . n

MISCELLANEOUS. 1 . 729, 3 69, 1 , 4 1 .2. 4 1x—5 1y .

3 . 9—30x + 37x2—20x3 + 4x4.

4 . 1 +x—x3

2x2+ 3x—444 0

91 m+ a. a .

3x _ 46 (4x

a8 3 9 10 £ 2, £ 23

733 7g 73 12. l . 1 362.

6 6 6’

3

14. —xy—2y2.

16 . (x (x

18. 5 . 1 9. 7. 20. £ 40.21 . 2a—2b—x—2y ,

23 . 514—31

4

( lb—b?

b2—4 a‘

3' 27. (16r

29. 1 43 , 5 233 . 30. 100.

(a?2—a2) (x

2—b? , (.v—a) —b).

ANSWERS . 323

23 4 . x—2

3 (4x — y)3 6 1 3 7. 4 . 38 2 3 9 30 1111111113638

4 0. 4 1 . 1 0x + 10z 4 2 7x2—2wy +y2,x

2 6m 7yz, 12x 4 1Ow3y x

2yz 205 9

3 l 2314.

l4 3 6—6 4 4 .

22_

a + 4 5w+ l

4 6 (a: 4 )(w 9)x2 + x + 2

1 6m um—1 48 1 °

25. 5 0 3 0 1bs

2 25 7 5 8

am +bu bm — an

6 1 l+1

1.

w1 1x + 2

4 4 .7m4 + 7m+ 2'

6 I 6 5’ 3

66 2 ; 4 .

69 2

70. 20 ; 4 0 years. 71 . 1 .

(x—2) (x (x—1 ) (x 74 . 0.

. 42. 76.

17, 77. 3 shflhngs, 2 shflhngs.

5 3 3

3w2 - x 4- 1 82. (mg

85 . 4 7 o r 74

4x 3—3xyz—i—5 y 3 88 5 25

92 —y4) 93 94 1 95 . 4

, 5 , 6

96 3 , 97. 20 miles. 98 Present price 3 pen ce

per dozen

83 . 3

86. 4 5 gallons.

89.

"

91 . x -

y .

mM- Byz

°

x4 —y4

324 ANSWERS .

101 . .6-4 — 1

,1 + xi + xi 102. (x

2- a

6). 1 03 . a .

104 . 1 . 105 . 1 3,4

. $ 106 .4 3 ;

o r $ 3 ; $ 5 ; $ 4 . 107. 20 shill ings .

108. 4 8. 1 1 1 .

294- 5 33 -5 10 6

7. 1 1 3 . 7 or

1 1 1 l1 15 . $2 ; $ 1 .

3,2;2,3

.

g._ x

2. 1 190 6 120

2 55 962 ab3 3 ab b29

2abz—b3 + a - b122.

—5 .vy + 2y

123 .

126. 5 , 127. 1 1 29. 3 .

130. 132.-4

13 3 . 4,—3 . 1 34 . x

2 13 5 . x4z a

‘o r

1 3 6 . $ 2. 1 37. 81 9615 . 138. 7—2 ,J3 .

a+ b—2a (a + c) (a+ b—2a)b _ a 2 (b _ a)

1 4 1 .

142. 1 97,—25 3—5x — 3 . 143 .

144 . (2) (3) 5 ; 7. 1 45 . (1) 3 , 11)(2) 8.

4 5 . 1 4 6 . 147. 20. 1 48. 1 6,

14 9.

4169. 1 5 0. As 5 to l . 1 5 1 . 1 5 2. a

’+ 4b.

1 5 3 . .6 —3 . 1 5 4 . (1) 5 . (2) 3 . (3 ) 7 ; 4 .

1 5 5 . (1) 8 . (2) 9. (3 ) $ 9 ; $ 7. 1 5 6 . 3 0 pence.

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326 ANSWERS

205 . (1) 7, £35" (2) l , —4 . (3 ) 4 3 ; $2. 206. 10miles.

207. 24 . 208. 6 crown s+ 18 shillings.209. 2x2+ 2ax + 4bs. 2 10. 7, 1 1 , 1 5 ,

g

21 1 .—2z2y + 3wy

2— 5 y3.

x1 2 + 5 x—282 2

'

4x2—25x - 1- 3 74 .

w3— l oz2 + 3 lx —3 0214 (I) 9

(3 ) 3 ; 215 . (1) 3 , - 6 . (2) 4 7 ; 4 5 . (3) 2, 4 ; 4 2 .

216 . 1 14 o f each . 217. 126 . 218 . 21 . 21 9.

1 3 , 14 . 220. 3 + 2 ,J2. 221 . x6+ x

3 + 1 , pza+ gx

—r .

arm - 1

a + b+ c

a—b—c'

224 . (1 ) 9. 225 .

—3 .

— 200. (3 )

223 .- 4 ) (3x — 2) (az

z

o f a mile . 27. 5 00 ; 1000 ; 4000. 228. 2hours ;

4 hours. 229. 2,5,8, 230. 12

2

(9

b (a2 b?) a

3 b32 I. 2a+2b+2c3 x

a (453 b2) (a b)

2(a2 62)

233 . (1) g. (2) 4

b—c * d234 . (1) 5 , (3 ) 5 ; 4 4 . 23 5 . 19 .

236 . 1 5 0, 5 0. 238. 1975 .

239. a + 2b+ 3a 240. xfiyjf + 8fi y ii

o 2_

24 1 1 4332. 242. mm . 243 . 105 shil

24 4 . 5 4 24 5 . 3 , 5 , 8 . 246. 6%per cent.

247. 2200. 248 . 249. 25 1 . 2a—b.

ANSWERS . 3 27

25 2. a“ 25 3 . 25 4 . (1) 5 .

(2) (3) 0, 25 5 : 1 12 ; 96 . 2 5 6 . A has

£ 5 400 E bas £ 7200. 25 7. 7; 1 3 . 25 8. 80.

260. £ 80. 261 . 62+ 2bc.

262. (510— 1 , (2x

‘+ 3ax3 — 4 2

2972

.2x2— 2x + 3 3263 . a. 2 4 1 +

x+x

. “

252+ 5 x—3 ’

1 . 260 .

6(2) 1 . (3) 18 ; 9. 266 . (1) 3 , —2. (2) 5 , 5

. (3 ) 2, 3 ; 3 , 2 .

268. 4 5 sh ilhngs, 30 shillings. 270. x2+§—1 ,

271. 0. 272. x2+ 3x + 8.

12a2 Saw4» 5 x2 4x4

a: (11 5 a2+ ax—2x2 (4 a

—3x ) (5 a—2x )’

az+ ax + x

2' 2 l 6 ° (1 ) 2°

d (a - l—b) d (a—b)277. (1)

a—b a + b(2) 4 , 7.

278 . m iles. 28 1 . 282. 2 (a — b) (o—J) ,

- 2bc. 284 . (1) 4 . (2) 6 ; 4 .

285 . (1) _ a’a,

286 . Second boat 1 6 minutes. 287. 3 feet ;

2 feet. 289 . 18 feet. 7} 2 01—1 )2 1 + x

291 . 0. 292. b2

X

295 .4

298 .

$4 4 1 5 : 1 11.

(2) 6 1 ; 73 . (3) 16 ; 8 . 296' (1) 7a_ 8 .

297. 144 minutes.

(3 ) l , 5 .

hours 771111 the st1 earn , 75 hours agamst the

l299 a —é

b, a , a +éb 300. 3 ,—1